2023. Number of Pairs of Strings With Concatenation Equal to Target

Problem

Given an array of digit strings nums and a digit string target, return **the number of pairs of indices *(i, j)* (where i != j) such that the concatenation of nums[i] + nums[j] equals **target.

  Example 1:

Input: nums = ["777","7","77","77"], target = "7777"
Output: 4
Explanation: Valid pairs are:
- (0, 1): "777" + "7"
- (1, 0): "7" + "777"
- (2, 3): "77" + "77"
- (3, 2): "77" + "77"

Example 2:

Input: nums = ["123","4","12","34"], target = "1234"
Output: 2
Explanation: Valid pairs are:
- (0, 1): "123" + "4"
- (2, 3): "12" + "34"

Example 3:

Input: nums = ["1","1","1"], target = "11"
Output: 6
Explanation: Valid pairs are:
- (0, 1): "1" + "1"
- (1, 0): "1" + "1"
- (0, 2): "1" + "1"
- (2, 0): "1" + "1"
- (1, 2): "1" + "1"
- (2, 1): "1" + "1"

  Constraints:

• 2 <= nums.length <= 100

• 1 <= nums[i].length <= 100

• 2 <= target.length <= 100

• nums[i] and target consist of digits.

• nums[i] and target do not have leading zeros.

Solution (Java)

class Solution {
    public int numOfPairs(String[] nums, String target) {
        int ans = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < nums.length; j++) {
                if (i != j) {
                    String con = nums[i] + nums[j];
                    if (con.equals(target)) {
                        ans++;
                    }
                }
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).