3356. Zero Array Transformation II

Description

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 amount by which each value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence, nums becomes a Zero Array. If no such k exists, return -1.

 

Example 1:

Input: nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]]

Output: 2

Explanation:

• For i = 0 (l = 0, r = 2, val = 1):
  • Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
  • The array will become [1, 0, 1].
• For i = 1 (l = 0, r = 2, val = 1):
  • Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
  • The array will become [0, 0, 0], which is a Zero Array. Therefore, the minimum value of k is 2.

Example 2:

Input: nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]]

Output: -1

Explanation:

• For i = 0 (l = 1, r = 3, val = 2):
  • Decrement values at indices [1, 2, 3] by [2, 2, 1] respectively.
  • The array will become [4, 1, 0, 0].
• For i = 1 (l = 0, r = 2, val= 1):
  • Decrement values at indices [0, 1, 2] by [1, 1, 0] respectively.
  • The array will become [3, 0, 0, 0], which is not a Zero Array.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 5 * 105
• 1 <= queries.length <= 105
• queries[i].length == 3
• 0 <= li <= ri < nums.length
• 1 <= vali <= 5

 

Solutions

Solution: Binary Search + Difference Array

• Time complexity: O(nlog(n+queries.length))
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number[][]} queries
 * @return {number}
 */
const minZeroArray = function (nums, queries) {
  if (nums.every(num => !num)) return 0;
  const n = nums.length;
  let left = 1;
  let right = queries.length;
  let result = -1;

  const transformToZeroArray = k => {
    const decrements = new Array(n + 1).fill(0);

    for (let index = 0; index < k; index++) {
      const [l, r, val] = queries[index];

      decrements[l] += val;
      decrements[r + 1] -= val;
    }

    for (let index = 1; index <= n; index++) {
      const num = nums[index - 1];

      if (num > decrements[index - 1]) return false;

      decrements[index] += decrements[index - 1];
    }

    return true;
  };

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    const isZeroArray = transformToZeroArray(mid);

    if (isZeroArray) {
      result = mid;
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }

  return result;
};