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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];
        for (var q : queries) {
            int l = q[0], r = q[1];
            ++d[l];
            --d[r + 1];
        }
        for (int i = 0, s = 0; i < n; ++i) {
            s += d[i];
            if (nums[i] > s) {
                return false;
            }
        }
        return true;
    }
}

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class Solution {
public:
    bool isZeroArray(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        int d[n + 1];
        memset(d, 0, sizeof(d));
        for (const auto& q : queries) {
            int l = q[0], r = q[1];
            ++d[l];
            --d[r + 1];
        }
        for (int i = 0, s = 0; i < n; ++i) {
            s += d[i];
            if (nums[i] > s) {
                return false;
            }
        }
        return true;
    }
};

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func isZeroArray(nums []int, queries [][]int) bool {
    d := make([]int, len(nums)+1)
    for _, q := range queries {
        l, r := q[0], q[1]
        d[l]++
        d[r+1]--
    }
    s := 0
    for i, x := range nums {
        s += d[i]
        if x > s {
            return false
        }
    }
    return true
}

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function isZeroArray(nums: number[], queries: number[][]): boolean {
    const n = nums.length;
    const d: number[] = Array(n + 1).fill(0);
    for (const [l, r] of queries) {
        ++d[l];
        --d[r + 1];
    }
    for (let i = 0, s = 0; i < n; ++i) {
        s += d[i];
        if (nums[i] > s) {
            return false;
        }
    }
    return true;
}

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