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Weekly Contest 424 Q2
Array
Prefix Sum

3355. Zero Array Transformation I

中文文档

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: 
    <ul>
	<li>Select the subset of indices as <code>[0, 2]</code> and decrement the values at these indices by 1.</li>
	<li>The array will become <code>[0, 0, 0]</code>, which is a Zero Array.</li>
</ul>
</li>

Example 2:

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

Output: false

Explanation:

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

 

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 $\textit{nums}[i] &gt; s$, 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.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}