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3355. Zero Array Transformation I

Description

You are given an integer array nums of length n and a 2D array queries, where queries[i] = [li, ri].

For each queries[i]:

  • Select a subset of indices within the range [li, ri] in nums.
  • Decrement the values at the selected indices by 1.

A Zero Array is an array where all elements are equal to 0.

Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false.

 

Example 1:

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

Output: true

Explanation:

  • For i = 0:
    • Select the subset of indices as [0, 2] and decrement the values at these indices by 1.
    • The array will become [0, 0, 0], which is a Zero Array.

Example 2:

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

Output: false

Explanation:

  • For i = 0:
    • Select the subset of indices as [1, 2, 3] and decrement the values at these indices by 1.
    • The array will become [4, 2, 1, 0].
  • For i = 1:
    • Select the subset of indices as [0, 1, 2] and decrement the values at these indices by 1.
    • The array will become [3, 1, 0, 0], which is not a Zero Array.

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 105
  • 1 <= queries.length <= 105
  • queries[i].length == 2
  • 0 <= li <= ri < nums.length

Solutions

Solution 1: Difference Array

We can use a difference array to solve this problem.

Define an array $d$ of length $n + 1$, with all initial values set to $0$. For each query $[l, r]$, we add $1$ to $d[l]$ and subtract $1$ from $d[r + 1]$.

Then we traverse the array $d$ within the range $[0, n - 1]$, accumulating the prefix sum $s$. If $\textit{nums}[i] > s$, it means $\textit{nums}$ cannot be converted to a zero array, so we return $\textit{false}$.

After traversing, return $\textit{true}$.

The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the array $\textit{nums}$ and the number of queries, respectively.

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class Solution:
    def isZeroArray(self, nums: List[int], queries: List[List[int]]) -> bool:
        d = [0] * (len(nums) + 1)
        for l, r in queries:
            d[l] += 1
            d[r + 1] -= 1
        s = 0
        for x, y in zip(nums, d):
            s += y
            if x > s:
                return False
        return True

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class Solution {
    public boolean isZeroArray(int[] nums, int[][] queries) {
        int n = nums.length;
        int[] d = new int[n + 1];