Number of Pairs of Strings With Concatenation Equal to Target

Problem Description

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.

Examples

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.

Approach

To solve this problem:

1. Initialize a counter count to keep track of the number of valid pairs.
2. Use nested loops to iterate through all pairs (i, j) such that i != j.
3. For each pair, check if the concatenation of nums[i] and nums[j] equals target.
   • If true, increment the counter count.
4. Return the counter count as the result.

This approach ensures that we check all possible pairs in the array, resulting in a time complexity of O(n^2) where n is the length of the nums array.

C++ Solution

class Solution {
public:
    int numOfPairs(vector<string>& nums, string target) {
        int count = 0;
        
        for (int i = 0; i < nums.size(); ++i) {
            for (int j = 0; j < nums.size(); ++j) {
                if (i != j && nums[i] + nums[j] == target) {
                    count++;
                }
            }
        }
        
        return count;
    }
};