You are given an integer array nums of length n and a 2D array queries where queries[i] = [li, ri, vali]. Each queries[i] represents the following action on nums: Decrement the value at each index in the range [li, ri] in nums by at most vali. The

minZeroArray.js

/**
 * Checks if the array can be made zero with a given number of queries.
 * @param {number[]} nums - The array of numbers.
 * @param {number[][]} queries - The array of queries.
 * @param {number} limit - The number of queries to use.
 * @return {boolean} - Returns true if the array can be made zero, false otherwise.
 */
let check = (nums, queries, limit) => {
    let arr = new Array(nums.length + 1).fill(0);

    // Apply each query to the arr array
    for (let i = 0; i < limit; i++) {
        arr[queries[i][0]] -= queries[i][2];
        arr[queries[i][1] + 1] += queries[i][2];
    }

    let dec = 0;
    let sum = 0;

    // Calculate the decayed values and sum up if they are positive
    for (let i = 0; i < nums.length; i++) {
        dec += arr[i];
        if (nums[i] + dec > 0) {
            sum += nums[i];
        }
    }

    // Return true if sum of positive elements is zero
    return sum == 0;
};

/**
 * Finds the minimum number of queries needed to make the array sum zero.
 * @param {number[]} nums - The array of numbers.
 * @param {number[][]} queries - The array of queries.
 * @return {number} - The minimum number of queries needed, or -1 if not possible.
 */
var minZeroArray = function(nums, queries) {
    let sum = 0;

    // Calculate the initial sum of the array
    for (let i = 0; i < nums.length; i++) {
        sum += nums[i];
    }

    // If the sum is already zero, no queries are needed
    if (sum == 0) {
        return 0;
    }

    let n = nums.length;
    let l = 1, r = queries.length;
    let res = -1;

    // Binary search to find the minimum number of queries needed
    while (l <= r) {
        let mid = Math.floor((l + r) / 2);

        if (check(nums, queries, mid)) {
            res = mid;
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }

    return res;
};