LeetCode in Kotlin

1658. Minimum Operations to Reduce X to Zero

Medium

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it is possible__, otherwise, return -1.

Example 1:

Input: nums = [1,1,4,2,3], x = 5

Output: 2

Explanation: The optimal solution is to remove the last two elements to reduce x to zero.

Example 2:

Input: nums = [5,6,7,8,9], x = 4

Output: -1

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10

Output: 5

Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104
• 1 <= x <= 109

Solution

class Solution {
    fun minOperations(nums: IntArray, x: Int): Int {
        var totalArraySum = 0
        for (each in nums) {
            totalArraySum += each
        }
        if (totalArraySum == x) {
            return nums.size
        }
        val target = totalArraySum - x
        // as we need to find value equal to x so that x-x=0,
        // and we need to search the longest sub array with sum equal t0 total array sum -x;
        var sum = 0
        var result = -1
        var start = 0
        for (end in nums.indices) {
            sum += nums[end]
            while (sum > target && start < nums.size) {
                sum -= nums[start]
                start++
            }
            if (sum == target) {
                result = Math.max(result, end + 1 - start)
            }
        }
        return if (result == -1) {
            result
        } else {
            nums.size - result
        }
    }
}

This site is open source. Improve this page.