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leetcode_java/src/main/java/LeetCodeJava/Greedy/MaximumSubarray.java
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| package LeetCodeJava.Greedy; | ||
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| // https://leetcode.com/problems/maximum-subarray/ | ||
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| public class MaximumSubarray { | ||
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| // V0 | ||
| // IDEA : BRUTE FORCE | ||
| // https://www.bilibili.com/video/BV1aY4y1Z7ya/?share_source=copy_web&vd_source=49348a1975630e1327042905e52f488a | ||
| public int maxSubArray(int[] nums) { | ||
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| if (nums.length == 0 || nums == null){ | ||
| return 0; | ||
| } | ||
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| if (nums.length == 1){ | ||
| return nums[0]; | ||
| } | ||
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| // loop over nums, reset start-idx if "current sub array" < current num | ||
| // maintain the max value (ans) as well while looping | ||
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| int ans = -Integer.MIN_VALUE; | ||
| int cumSum = 0; | ||
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| for (int i = 0; i < nums.length; i++){ | ||
| if (cumSum + nums[i] < 0){ | ||
| ans = Math.max(ans, cumSum + nums[i]); | ||
| cumSum = 0; | ||
| }else{ | ||
| cumSum += nums[i]; | ||
| ans = Math.max(ans, cumSum); | ||
| } | ||
| } | ||
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| return ans; | ||
| } | ||
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| // V1 | ||
| // IDEA : OPTIMIZED BRUTE FORCE (TLE) | ||
| // https://leetcode.com/problems/maximum-subarray/editorial/ | ||
| public int maxSubArray_2(int[] nums) { | ||
| int maxSubarray = Integer.MIN_VALUE; | ||
| for (int i = 0; i < nums.length; i++) { | ||
| int currentSubarray = 0; | ||
| for (int j = i; j < nums.length; j++) { | ||
| currentSubarray += nums[j]; | ||
| maxSubarray = Math.max(maxSubarray, currentSubarray); | ||
| } | ||
| } | ||
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| return maxSubarray; | ||
| } | ||
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| // V2 | ||
| // IDEA : Dynamic Programming, Kadane's Algorithm | ||
| // https://leetcode.com/problems/maximum-subarray/editorial/ | ||
| public int maxSubArray_3(int[] nums) { | ||
| // Initialize our variables using the first element. | ||
| int currentSubarray = nums[0]; | ||
| int maxSubarray = nums[0]; | ||
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| // Start with the 2nd element since we already used the first one. | ||
| for (int i = 1; i < nums.length; i++) { | ||
| int num = nums[i]; | ||
| // If current_subarray is negative, throw it away. Otherwise, keep adding to it. | ||
| currentSubarray = Math.max(num, currentSubarray + num); | ||
| maxSubarray = Math.max(maxSubarray, currentSubarray); | ||
| } | ||
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| return maxSubarray; | ||
| } | ||
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| // V3 | ||
| // IDEA : Divide and Conquer | ||
| // https://leetcode.com/problems/maximum-subarray/editorial/ | ||
| private int[] numsArray; | ||
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| public int maxSubArray_4(int[] nums) { | ||
| numsArray = nums; | ||
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| // Our helper function is designed to solve this problem for | ||
| // any array - so just call it using the entire input! | ||
| return findBestSubarray(0, numsArray.length - 1); | ||
| } | ||
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| private int findBestSubarray(int left, int right) { | ||
| // Base case - empty array. | ||
| if (left > right) { | ||
| return Integer.MIN_VALUE; | ||
| } | ||
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| int mid = Math.floorDiv(left + right, 2); | ||
| int curr = 0; | ||
| int bestLeftSum = 0; | ||
| int bestRightSum = 0; | ||
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| // Iterate from the middle to the beginning. | ||
| for (int i = mid - 1; i >= left; i--) { | ||
| curr += numsArray[i]; | ||
| bestLeftSum = Math.max(bestLeftSum, curr); | ||
| } | ||
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| // Reset curr and iterate from the middle to the end. | ||
| curr = 0; | ||
| for (int i = mid + 1; i <= right; i++) { | ||
| curr += numsArray[i]; | ||
| bestRightSum = Math.max(bestRightSum, curr); | ||
| } | ||
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| // The bestCombinedSum uses the middle element and the best | ||
| // possible sum from each half. | ||
| int bestCombinedSum = numsArray[mid] + bestLeftSum + bestRightSum; | ||
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| // Find the best subarray possible from both halves. | ||
| int leftHalf = findBestSubarray(left, mid - 1); | ||
| int rightHalf = findBestSubarray(mid + 1, right); | ||
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| // The largest of the 3 is the answer for any given input array. | ||
| return Math.max(bestCombinedSum, Math.max(leftHalf, rightHalf)); | ||
| } | ||
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| } |