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3643. Zero Array Transformation Ii

3643. Zero Array Transformation II

Medium

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

Solution

class Solution {
    public int minZeroArray(int[] nums, int[][] queries) {
        int n = nums.length;
        int low = 0, high = queries.length - 1, ans = -1;
        boolean flag = true;
        for (int ele : nums) if (ele != 0) flag = false;
        if (flag == true) return 0;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (ok(mid, queries, nums)) {
                ans = mid;
                high = mid - 1;
            }
            else low = mid + 1;
        }
        if (ans == -1) return -1;
        return ans + 1;
    }

    private boolean ok(int mid, int[][] queries, int nums[]) {
        int n = nums.length;
        int arr[] = new int[n];
        for (int i = 0; i <= mid; i++) {
            int l = queries[i][0], r = queries[i][1], val = queries[i][2];
            arr[l] += val;
            if (r + 1 < n) arr[r + 1] -= val;
        }

        for (int i = 1; i < n; i++) arr[i] += arr[i - 1];
        for (int i = 0; i < n; i++) {
            if (nums[i] > arr[i]) return false;
        }
        return true;
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Explanation

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