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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0053_maximum_subarray;

// #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Array #Dynamic_Programming
// #Divide_and_Conquer #Data_Structure_I_Day_1_Array #Dynamic_Programming_I_Day_5
// #Udemy_Famous_Algorithm #Big_O_Time_O(n)_Space_O(1)
// #2023_08_11_Time_1_ms_(100.00%)_Space_57.7_MB_(90.58%)

/**
 * 53 - Maximum Subarray\.
 *
 * Easy
 *
 * Given an integer array `nums`, find the contiguous subarray (containing at least one number) which has the largest sum and return _its sum_.
 *
 * A **subarray** is a **contiguous** part of an array.
 *
 * **Example 1:**
 *
 * **Input:** nums = [-2,1,-3,4,-1,2,1,-5,4]
 *
 * **Output:** 6
 *
 * **Explanation:** [4,-1,2,1] has the largest sum = 6. 
 *
 * **Example 2:**
 *
 * **Input:** nums = [1]
 *
 * **Output:** 1 
 *
 * **Example 3:**
 *
 * **Input:** nums = [5,4,-1,7,8]
 *
 * **Output:** 23 
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   -10⁴ <= nums[i] <= 10⁴
 *
 * **Follow up:** If you have figured out the `O(n)` solution, try coding another solution using the **divide and conquer** approach, which is more subtle.
**/
public class Solution {
    public int maxSubArray(int[] nums) {
        int maxi = Integer.MIN_VALUE;
        int sum = 0;
        for (int num : nums) {
            // calculating sub-array sum
            sum += num;
            maxi = Math.max(sum, maxi);
            if (sum < 0) {
                // there is no point to carry a -ve subarray sum. hence setting to 0
                sum = 0;
            }
        }
        return maxi;
    }
}