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"

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:

from collections import Counter

class Solution:
    def numOfPairs(self, nums: List[str], target: str) -> int:
        k = len(target)
        nums = Counter(nums)
        ctr = 0
        for num in nums:
            if target.startswith(num):
                chunk = target[len(num) - k:]
                if chunk in nums:
                    ctr += nums[num] * nums[chunk]
                    if num == chunk:
                        ctr -= nums[chunk]
        return ctr