1497. Check If Array Pairs Are Divisible by k.md

1497. Check If Array Pairs Are Divisible by k

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the second one in Weekly Contest 195.

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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 <= 10^5
• n is even.
• -10^9 <= arr[i] <= 10^9
• 1 <= k <= 10^5

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Solution

• mine
  • Java

    • Time Limit Exceeded

      // O(N^2)time O(N)space
      public boolean canArrange(int[] arr, int k) {
          int len = arr.length;
          boolean[] t = new boolean[len];
          for(int i = 0; i < len; i++){
              if(t[i]){
                  continue;
              }
              for(int j = i + 1; j < len; j++){
                  if(t[j]){
                      continue;
                  }
                  if((arr[i] + arr[j]) % k == 0){
                      t[i] = true;
                      t[j] = true;
                      break;
                  }
              }
          }
          boolean res = true;
          for(boolean b : t){
              res &= b;
          }
          return res;
      }
      

    • Runtime: 10 ms, faster than 64.41%, Memory Usage: 79.5 MB, less than 100.00% of Java online submissions

      // O(M)time O(K)space
      // M = max(arr.length, k/2)
      public boolean canArrange(int[] arr, int k) {
          int len = arr.length;
          int[] record = new int[k];
          for(int i = 0; i < len; i++){
              int mod = (arr[i] % k + k) % k;
              record[mod]++;
          }
          if(record[0] % 2 != 0){
              return false;
          }
          for(int i = 1; i <= k / 2; i++){
              if(record[i] != record[k - i]){
                  return false;
              }
          }
          return true;
      }
      

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