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...c/main/java/LeetCodeJava/BinarySearch/FindFirstAndLastPositionOfElementInSortedArray.java
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| package LeetCodeJava.BinarySearch; | ||
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| // https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/ | ||
| /** | ||
| * 34. Find First and Last Position of Element in Sorted Array | ||
| * Solved | ||
| * Medium | ||
| * Topics | ||
| * Companies | ||
| * Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. | ||
| * | ||
| * If target is not found in the array, return [-1, -1]. | ||
| * | ||
| * You must write an algorithm with O(log n) runtime complexity. | ||
| * | ||
| * | ||
| * | ||
| * Example 1: | ||
| * | ||
| * Input: nums = [5,7,7,8,8,10], target = 8 | ||
| * Output: [3,4] | ||
| * Example 2: | ||
| * | ||
| * Input: nums = [5,7,7,8,8,10], target = 6 | ||
| * Output: [-1,-1] | ||
| * Example 3: | ||
| * | ||
| * Input: nums = [], target = 0 | ||
| * Output: [-1,-1] | ||
| * | ||
| * | ||
| * Constraints: | ||
| * | ||
| * 0 <= nums.length <= 105 | ||
| * -109 <= nums[i] <= 109 | ||
| * nums is a non-decreasing array. | ||
| * -109 <= target <= 109 | ||
| * | ||
| */ | ||
| public class FindFirstAndLastPositionOfElementInSortedArray { | ||
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| // V0 | ||
| // TODO : implement | ||
| // public int[] searchRange(int[] nums, int target) { | ||
| // | ||
| // } | ||
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| // V1 | ||
| // IDEA : BINARY SEARCH | ||
| // https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/solutions/14734/easy-java-o-logn-solution/ | ||
| public int[] searchRange_1(int[] nums, int target) { | ||
| int[] result = new int[2]; | ||
| result[0] = findFirst(nums, target); | ||
| result[1] = findLast(nums, target); | ||
| return result; | ||
| } | ||
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| private int findFirst(int[] nums, int target) { | ||
| int idx = -1; | ||
| int start = 0; | ||
| int end = nums.length - 1; | ||
| while (start <= end) { | ||
| int mid = (start + end) / 2; | ||
| if (nums[mid] >= target) { | ||
| end = mid - 1; | ||
| } else { | ||
| start = mid + 1; | ||
| } | ||
| if (nums[mid] == target) | ||
| idx = mid; | ||
| } | ||
| return idx; | ||
| } | ||
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| private int findLast(int[] nums, int target) { | ||
| int idx = -1; | ||
| int start = 0; | ||
| int end = nums.length - 1; | ||
| while (start <= end) { | ||
| int mid = (start + end) / 2; | ||
| if (nums[mid] <= target) { | ||
| start = mid + 1; | ||
| } else { | ||
| end = mid - 1; | ||
| } | ||
| if (nums[mid] == target) | ||
| idx = mid; | ||
| } | ||
| return idx; | ||
| } | ||
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| // V2 | ||
| // IDEA : Binary Tree | ||
| // https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/solutions/14701/a-very-simple-java-solution-with-only-one-binary-search-algorithm/ | ||
| public int[] searchRange_2(int[] A, int target) { | ||
| int start = firstGreaterEqual(A, target); | ||
| if (start == A.length || A[start] != target) { | ||
| return new int[]{-1, -1}; | ||
| } | ||
| return new int[]{start, firstGreaterEqual(A, target + 1) - 1}; | ||
| } | ||
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| //find the first number that is greater than or equal to target. | ||
| //could return A.length if target is greater than A[A.length-1]. | ||
| //actually this is the same as lower_bound in C++ STL. | ||
| private static int firstGreaterEqual(int[] A, int target) { | ||
| int low = 0, high = A.length; | ||
| while (low < high) { | ||
| int mid = low + ((high - low) >> 1); | ||
| //low <= mid < high | ||
| if (A[mid] < target) { | ||
| low = mid + 1; | ||
| } else { | ||
| //should not be mid-1 when A[mid]==target. | ||
| //could be mid even if A[mid]>target because mid<high. | ||
| high = mid; | ||
| } | ||
| } | ||
| return low; | ||
| } | ||
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| // V3 | ||
| // IDEA : binary Search | ||
| // https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/solutions/5378191/video-binary-search-solution/ | ||
| public int[] searchRange_3(int[] nums, int target) { | ||
| int[] result = { -1, -1 }; | ||
| int left = binarySearch(nums, target, true); | ||
| int right = binarySearch(nums, target, false); | ||
| result[0] = left; | ||
| result[1] = right; | ||
| return result; | ||
| } | ||
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| private int binarySearch(int[] nums, int target, boolean isSearchingLeft) { | ||
| int left = 0; | ||
| int right = nums.length - 1; | ||
| int idx = -1; | ||
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| while (left <= right) { | ||
| int mid = left + (right - left) / 2; | ||
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| if (nums[mid] > target) { | ||
| right = mid - 1; | ||
| } else if (nums[mid] < target) { | ||
| left = mid + 1; | ||
| } else { | ||
| idx = mid; | ||
| if (isSearchingLeft) { | ||
| right = mid - 1; | ||
| } else { | ||
| left = mid + 1; | ||
| } | ||
| } | ||
| } | ||
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| return idx; | ||
| } | ||
| } |
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