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leetcode_java/src/main/java/LeetCodeJava/HashTable/ContinuousSubarraySum.java
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| package LeetCodeJava.HashTable; | ||
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| // https://leetcode.com/problems/continuous-subarray-sum/description/ | ||
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| import java.util.HashMap; | ||
| import java.util.Map; | ||
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| /** | ||
| * 523. Continuous Subarray Sum | ||
| * Solved | ||
| * Medium | ||
| * Topics | ||
| * Companies | ||
| * Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise. | ||
| * | ||
| * A good subarray is a subarray where: | ||
| * | ||
| * its length is at least two, and | ||
| * the sum of the elements of the subarray is a multiple of k. | ||
| * Note that: | ||
| * | ||
| * A subarray is a contiguous part of the array. | ||
| * An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k. | ||
| * | ||
| * | ||
| * Example 1: | ||
| * | ||
| * Input: nums = [23,2,4,6,7], k = 6 | ||
| * Output: true | ||
| * Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6. | ||
| * Example 2: | ||
| * | ||
| * Input: nums = [23,2,6,4,7], k = 6 | ||
| * Output: true | ||
| * Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42. | ||
| * 42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer. | ||
| * Example 3: | ||
| * | ||
| * Input: nums = [23,2,6,4,7], k = 13 | ||
| * Output: false | ||
| * | ||
| * | ||
| * Constraints: | ||
| * | ||
| * 1 <= nums.length <= 105 | ||
| * 0 <= nums[i] <= 109 | ||
| * 0 <= sum(nums[i]) <= 231 - 1 | ||
| * 1 <= k <= 231 - 1 | ||
| * Seen this question in a real interview before? | ||
| * 1/5 | ||
| * Yes | ||
| * No | ||
| * Accepted | ||
| * 625K | ||
| * Submissions | ||
| * 2M | ||
| * Acceptance Rate | ||
| * 30.5% | ||
| */ | ||
| public class ContinuousSubarraySum { | ||
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| // V0 | ||
| // IDEA : HASH MAP (fixed by gpt) | ||
| public boolean checkSubarraySum(int[] nums, int k) { | ||
| if (nums.length < 2) { | ||
| return false; | ||
| } | ||
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| // Map to store the remainder and its index | ||
| Map<Integer, Integer> map = new HashMap<>(); | ||
| // Initialize with remainder 0 at index -1 to handle edge cases | ||
| map.put(0, -1); | ||
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| int presum = 0; | ||
| for (int i = 0; i < nums.length; i++) { | ||
| presum += nums[i]; | ||
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| // Calculate remainder | ||
| int remainder = (k != 0) ? presum % k : presum; | ||
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| // If the remainder already exists in the map | ||
| if (map.containsKey(remainder)) { | ||
| // Check if the subarray length is at least 2 | ||
| if (i - map.get(remainder) > 1) { | ||
| return true; | ||
| } | ||
| } else { | ||
| // Store the first occurrence of this remainder | ||
| map.put(remainder, i); | ||
| } | ||
| } | ||
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| return false; | ||
| } | ||
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| // V1 | ||
| // IDEA : HASHMAP | ||
| // https://leetcode.com/problems/continuous-subarray-sum/editorial/ | ||
| public boolean checkSubarraySum_1(int[] nums, int k) { | ||
| int prefixMod = 0; | ||
| HashMap<Integer, Integer> modSeen = new HashMap<>(); | ||
| modSeen.put(0, -1); | ||
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| for (int i = 0; i < nums.length; i++) { | ||
| /** | ||
| * NOTE !!! we get `mod of prefixSum`, instead of get prefixSum | ||
| */ | ||
| prefixMod = (prefixMod + nums[i]) % k; | ||
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| if (modSeen.containsKey(prefixMod)) { | ||
| // ensures that the size of subarray is at least 2 | ||
| if (i - modSeen.get(prefixMod) > 1) { | ||
| return true; | ||
| } | ||
| } else { | ||
| // mark the value of prefixMod with the current index. | ||
| modSeen.put(prefixMod, i); | ||
| } | ||
| } | ||
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| return false; | ||
| } | ||
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| // V2 | ||
| // IDEA : HASHMAP | ||
| // https://leetcode.com/problems/continuous-subarray-sum/solutions/1405425/a-java-implementation-of-prefix-sum-that-si0m/ | ||
| public boolean checkSubarraySum_2(int[] nums, int k) { | ||
| // maintain a hash map to store <sum % k, index> | ||
| Map<Integer, Integer> map = new HashMap<>(); | ||
| int sum = 0; | ||
| for (int i = 0; i < nums.length; i++) { | ||
| sum += nums[i]; | ||
| sum %= k; | ||
| // case 1 | ||
| if (sum == 0 && i > 0) { | ||
| return true; | ||
| } | ||
| // case 2 | ||
| if (map.containsKey(sum) && i - map.get(sum) > 1) { | ||
| return true; | ||
| } | ||
| if (!map.containsKey(sum)) { | ||
| map.put(sum, i); | ||
| } | ||
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| } | ||
| return false; | ||
| } | ||
| } |
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