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NeoYYNeoYY
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‎15. 3Sum.cpp

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/*
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Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
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Note:
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The solution set must not contain duplicate triplets.
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Example:
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Given array nums = [-1, 0, 1, 2, -1, -4],
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A solution set is:
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[
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[-1, 0, 1],
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[-1, -1, 2]
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]
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Sort first, then do comparison, early break, etc..
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*/
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class Solution {
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public:
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vector<vector<int>> threeSum(vector<int>& nums) {
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sort( nums.begin(), nums.end() );
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vector<vector<int>> res;
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if ( nums.empty() || nums.front() > 0 || nums.back() < 0 ) return res;
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for ( int i = 0; i <= (int)nums.size() - 3; ++i )
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{
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if ( nums[i] > 0 ) break;
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if ( i > 0 && nums[i] == nums[i - 1] ) continue;
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int target = -nums[i];
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int j = i + 1, k = nums.size() - 1;
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while ( j < k )
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{
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if ( nums[j] + nums[k] == target )
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{
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res.push_back( {nums[i], nums[j], nums[k]} );
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while ( j < k && nums[j + 1] == nums[j] ) ++j;
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while ( j < k && nums[k - 1] == nums[k] ) --k;
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++j;
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--k;
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}
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else if ( nums[j] + nums[k] < target )
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++j;
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else
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--k;
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}
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}
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return res;
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}
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};

‎16. 3Sum Closest.cpp

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/*
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Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
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Example 1:
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Input: nums = [-1,2,1,-4], target = 1
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Output: 2
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Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
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Constraints:
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3 <= nums.length <= 10^3
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-10^3 <= nums[i] <= 10^3
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-10^4 <= target <= 10^4
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Solution similar to the previous question.
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*/
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class Solution {
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public:
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int threeSumClosest(vector<int>& nums, int target) {
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sort( nums.begin(), nums.end() );
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int sum = nums[0] + nums[1] + nums[2], diff = abs( sum - target ), n = nums.size();
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if ( target > 0 && nums.back() < 0 ) return nums[n - 1] + nums[n - 2] + nums[n - 3];
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for ( int i = 0; i < n - 2; ++i )
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{
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if ( nums[i] * 3 > target ) return min( sum, nums[i] + nums[i + 1] + nums[i + 2] );
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int j = i + 1, k = n - 1;
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while ( j < k )
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{
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int tmp = nums[i] + nums[j] + nums[k];
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int newDiff = abs( tmp - target );
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if ( newDiff < diff )
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{
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diff = newDiff;
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sum = tmp;
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}
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if ( tmp < target ) ++j;
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else --k;
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}
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}
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return sum;
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}
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};
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//Optimized
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class Solution {
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public:
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int threeSumClosest(vector<int>& nums, int target) {
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sort( nums.begin(), nums.end() );
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int sum = nums[0] + nums[1] + nums[2], diff = abs( sum - target ), n = nums.size();
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if ( target > 0 && nums.back() < 0 ) return nums[n - 1] + nums[n - 2] + nums[n - 3];
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for ( int i = 0; i < n - 2; ++i )
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{
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if ( nums[i] * 3 > target ) return min( sum, nums[i] + nums[i + 1] + nums[i + 2] );
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if ( i > 0 && nums[i] == nums[i - 1] ) continue;
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int j = i + 1, k = n - 1;
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while ( j < k )
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{
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int tmp = nums[i] + nums[j] + nums[k];
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if ( tmp == target ) return tmp;
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int newDiff = abs( tmp - target );
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if ( newDiff < diff )
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{
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diff = newDiff;
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sum = tmp;
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}
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if ( tmp < target )
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{
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while ( ++j < k && nums[j] == nums[j - 1] );
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}
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else
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{
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while ( j < --k && nums[k] == nums[k + 1] );
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}
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}
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}
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return sum;
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}
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};

‎17. Letter Combinations of a Phone Number.cpp

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/*
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Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.
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A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
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Example:
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Input: "23"
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Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
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Note:
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Although the above answer is in lexicographical order, your answer could be in any order you want.
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Solution one is iterative,
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two is recursive
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*/
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class Solution {
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public:
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vector<string> letterCombinations(string digits) {
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if (digits.empty()) return {};
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vector<string> res{""};
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vector<string> dict = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
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for ( int i = 0; i < digits.size(); ++i )
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{
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vector<string> temp;
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string tp = dict[digits[i] - '0'];
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for ( int j = 0; j < tp.size(); ++j )
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for ( auto &a: res ) temp.push_back( a + tp[j] );
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res = temp;
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}
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return res;
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}
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};
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class Solution {
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public:
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vector<string> letterCombinations(string digits) {
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if (digits.empty()) return {};
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vector<string> res;
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vector<string> dict = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
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DFSCombine( dict, 0, "", res, digits );
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return res;
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}
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void DFSCombine( vector<string>& dict, int level, string out, vector<string>& res, string digits )
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{
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if ( level == digits.size() )
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{
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res.push_back( out );
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return;
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}
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string temp = dict[digits[level] - '0'];
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for ( int i = 0; i < temp.size(); ++i )
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{
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DFSCombine( dict, level + 1, out + temp[i], res, digits );
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}
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}
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};

‎18. 4Sum.cpp

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/*
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Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
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Note:
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The solution set must not contain duplicate quadruplets.
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Example:
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Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.
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A solution set is:
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[
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[-1, 0, 0, 1],
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[-2, -1, 1, 2],
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[-2, 0, 0, 2]
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]
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Solution one using two pointers, traditional Solution
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Two using the characteristic of set
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*/
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class Solution {
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public:
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vector<vector<int>> fourSum(vector<int>& nums, int target) {
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set<vector<int>> res;
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int n = nums.size();
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sort( nums.begin(), nums.end() );
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if ( nums.empty() || ( target < 0 && nums.front() > 0 ) || ( target > 0 && nums.back() < 0 ) ) return {};
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for ( int i = 0; i < n - 3; ++i )
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{
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for ( int j = i + 1; j < n - 2; ++j )
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{
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int k = j + 1, z = n - 1;
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while ( k < z )
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{
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int sum = nums[i] + nums[j] + nums[k] + nums[z];
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if ( target == sum )
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{
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vector<int> temp = { nums[i], nums[j], nums[k], nums[z] };
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res.insert( temp );
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++k; --z;
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}
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else if ( target < sum )
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{
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--z;
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}
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else ++k;
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}
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}
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}
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return vector<vector<int>>( res.begin(), res.end() );
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}
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};
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//Recursive
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class Solution {
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public:
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vector<vector<int>> fourSum(vector<int>& nums, int target) {
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set<vector<int>> res;
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int n = nums.size();
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sort( nums.begin(), nums.end() );
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if ( nums.empty() || ( target < 0 && nums.front() > 0 ) || ( target > 0 && nums.back() < 0 ) ) return {};
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for ( int i = 0; i < n - 3; ++i )
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{
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for ( int j = i + 1; j < n - 2; ++j )
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{
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int k = j + 1, z = n - 1;
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while ( k < z )
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{
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int sum = nums[i] + nums[j] + nums[k] + nums[z];
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if ( target == sum )
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{
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vector<int> temp = { nums[i], nums[j], nums[k], nums[z] };
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res.insert( temp );
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++k; --z;
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}
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else if ( target < sum )
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{
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--z;
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}
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else ++k;
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}
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}
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}
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return vector<vector<int>>( res.begin(), res.end() );
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}
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};

‎201. Bitwise AND of Numbers Range.cpp

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/*
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Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
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Example 1:
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Input: [5,7]
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Output: 4
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Example 2:
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Input: [0,1]
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Output: 0
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Solution one using mask
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Solution two shift m, n, until they are equal
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Solution three is recursive
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Solution four using n & ( n - 1 )
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*/
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class Solution {
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public:
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int rangeBitwiseAnd(int m, int n) {
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if ( m == 0 ) return 0;
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unsigned int d = 0x7fffffff;
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while ((m & d) != (n & d)) {
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d <<= 1;
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}
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return m & d;
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}
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};
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class Solution {
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public:
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int rangeBitwiseAnd(int m, int n) {
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if ( m == 0 || n == 0 ) return 0;
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int i = 0;
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while ( m != n )
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{
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m >>= 1;
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n >>= 1;
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i++;
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}
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return m << i;
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}
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};
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class Solution {
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public:
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int rangeBitwiseAnd(int m, int n) {
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return (n > m) ? (rangeBitwiseAnd(m / 2, n / 2) << 1) : m;
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}
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};
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class Solution {
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public:
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int rangeBitwiseAnd(int m, int n) {
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while ( n > m )
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{
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n = n & ( n - 1 );
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}
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return n;
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}
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};

‎303. Range Sum Query - Immutable.cpp

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/*
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Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
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Example:
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Given nums = [-2, 0, 3, -5, 2, -1]
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sumRange(0, 2) -> 1
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sumRange(2, 5) -> -1
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sumRange(0, 5) -> -3
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Note:
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You may assume that the array does not change.
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There are many calls to sumRange function.
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Solution is using dp
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*/
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class NumArray {
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public:
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NumArray(vector<int>& nums) {
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dp.resize( nums.size() + 1, 0 );
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for ( int i = 1; i <= nums.size(); ++i )
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{
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dp[i] = dp[i - 1] + nums[i - 1];
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}
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}
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int sumRange(int i, int j) {
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return dp[j + 1] - dp[i];
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}
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private:
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vector<int> dp;
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};
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/**
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* Your NumArray object will be instantiated and called as such:
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* NumArray* obj = new NumArray(nums);
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* int param_1 = obj->sumRange(i,j);
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*/

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