# [Programming Problem] Number of Pairs of Strings With Concatenation Equal to Target

```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"```

My initial thought was to use some sort of advanced pattern searching algorithm like Robin Karp. But there was a more straightforward way simply using hashtables (involing suffixes and prefixes).

– Add all suffixes and prefixes of target into a set
– Loop through nums and check if both suffix and prefix are part of the set.
Gotcha: In the example, target=’7777′ nums: [‘7777’, ‘777’], (0, 1) would show up as a pair which its clearly not.
– Add another condition to make sure length of nums[i] + nums[j] equals length of target.
Another gotcha was how we build suffixes (look at comment below).

```/** * @param {string[]} nums * @param {string} target * @return {number} */ var numOfPairs = function(nums, target) { const prefixes = new Set(); const suffixes = new Set();   let curr = ''; for ( let i = 0 ; i < target.length ; i++ ) { curr += target[i]; prefixes.add(curr); }   curr = ''; for ( let i = target.length-1 ; i >= 0 ; i-- ) { // curr += target[i] will fail. // If you do that, you'll be comparing suffixes in reverse order. curr = target[i] + curr; suffixes.add(curr); }   let cnt = 0; for ( let i = 0 ; i < nums.length ; i++ ) { for ( let j = 0 ; j < nums.length ; j++ ) { if ( i === j ) continue; if ( prefixes.has(nums[i]) && suffixes.has(nums[j]) && (nums[i].length + nums[j].length) === target.length ) { cnt++ } } }   return cnt; };```