g1701_1800.s1800_maximum_ascending_subarray_sum.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1701_1800.s1800_maximum_ascending_subarray_sum;

// #Easy #Array #2022_04_19_Time_0_ms_(100.00%)_Space_42.1_MB_(22.29%)

/**
 * 1800 - Maximum Ascending Subarray Sum\.
 *
 * Easy
 *
 * Given an array of positive integers `nums`, return the _maximum possible sum of an **ascending** subarray in_ `nums`.
 *
 * A subarray is defined as a contiguous sequence of numbers in an array.
 *
 * A subarray [nums_l, nums_l+1, ..., nums_r-1, nums_r] is **ascending** if for all `i` where `l <= i < r`, nums_i < nums_i+1. Note that a subarray of size `1` is **ascending**.
 *
 * **Example 1:**
 *
 * **Input:** nums = [10,20,30,5,10,50]
 *
 * **Output:** 65
 *
 * **Explanation:** [5,10,50] is the ascending subarray with the maximum sum of 65.
 *
 * **Example 2:**
 *
 * **Input:** nums = [10,20,30,40,50]
 *
 * **Output:** 150
 *
 * **Explanation:** [10,20,30,40,50] is the ascending subarray with the maximum sum of 150.
 *
 * **Example 3:**
 *
 * **Input:** nums = [12,17,15,13,10,11,12]
 *
 * **Output:** 33
 *
 * **Explanation:** [10,11,12] is the ascending subarray with the maximum sum of 33.
 *
 * **Constraints:**
 *
 * *   `1 <= nums.length <= 100`
 * *   `1 <= nums[i] <= 100`
**/
public class Solution {
    public int maxAscendingSum(int[] nums) {
        int maxSum = nums[0];
        int i = 0;
        int j = i + 1;
        while (i < nums.length - 1 && j < nums.length) {
            int sum = nums[j - 1];
            while (j < nums.length && nums[j] - nums[j - 1] > 0) {
                sum += nums[j];
                j++;
            }
            i = j;
            maxSum = Math.max(maxSum, sum);
            j++;
        }
        return maxSum;
    }
}