README_EN.md

1497. Check If Array Pairs Are Divisible by k

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Description

Given an array of integers arr of even length n and an integer k.

We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.

Return True If you can find a way to do that or False otherwise.

 

Example 1:

Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
Output: true
Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).

Example 2:

Input: arr = [1,2,3,4,5,6], k = 7
Output: true
Explanation: Pairs are (1,6),(2,5) and(3,4).

Example 3:

Input: arr = [1,2,3,4,5,6], k = 10
Output: false
Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.

Example 4:

Input: arr = [-10,10], k = 2
Output: true

Example 5:

Input: arr = [-1,1,-2,2,-3,3,-4,4], k = 3
Output: true

 

Constraints:

• arr.length == n
• 1 <= n <= 105
• n is even.
• -109 <= arr[i] <= 109
• 1 <= k <= 105

Solutions

Python3

class Solution:
    def isPathCrossing(self, path: str) -> bool:
        x = y = 0
        visited = set()
        visited.add('0.0')
        for c in path:
            if c == 'N':
                y += 1
            elif c == 'S':
                y -= 1
            elif c == 'E':
                x += 1
            else:
                x -= 1
            pos = f'{x}.{y}'
            if pos in visited:
                return True
            visited.add(pos)
        return False

Java

class Solution {
    public boolean isPathCrossing(String path) {
        Set<String> visited = new HashSet<>();
        visited.add("0.0");
        int x = 0, y = 0;
        for (int i = 0; i < path.length(); ++i) {
            char c = path.charAt(i);
            if (c == 'N') {
                ++y;
            } else if (c == 'S') {
                --y;
            } else if (c == 'E') {
                ++x;
            } else {
                --x;
            }
            String pos = x + "." + y;
            if (visited.contains(pos)) {
                return true;
            }
            visited.add(pos);
        }
        return false;
    }
}

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