Count Prefix And Suffix Pairs I

Intuition

The problem can be solved with a brute-force approach where we iterate over all possible pairs and check the condition for each.

Approach

Below is the step-by-step breakdown of the approach:

1. Helper Function:

   • Define a helper function isPrefixAndSuffix, using startsWith and endsWith to check if one string is both a prefix and a suffix of another string.
   • Return false early if the first string is longer than the second string, since it can't be a valid prefix or suffix.

2. Iterate Over Pairs:

   • Use two nested loops to iterate over all possible pairs of words in the array, ensuring  i < j  to avoid redundant checks.
   • For each pair, use the helper function to determine if the condition is met, and increment the count accordingly.

3. Return the Count:

   • After iterating through all pairs, return the total count of valid prefix-suffix pairs.

Complexity

• Time Complexity: $O(n^2\cdot L)$,
  • The brute-force approach involves  $O(n^2)$  pair comparisons, where  n  is the number of words.
  • For each pair, the isPrefixAndSuffix function has a worst-case complexity of  $O(L)$ , where  L  is the average length of the words.
• Space Complexity: $O(1)$ , as the function uses constant extra space.

Code

function countPrefixSuffixPairs(words: string[]): number {
  const isPrefixAndSuffix = (str1: string, str2: string): boolean => {
    if (str1.length > str2.length) return false
    return str2.startsWith(str1) && str2.endsWith(str1)
  }
 
  let count = 0
  const n = words.length
  for (let i = 0; i < n - 1; i++) {
    for (let j = i + 1; j < n; j++) {
      if (isPrefixAndSuffix(words[i], words[j])) {
        count++
      }
    }
  }
 
  return count
}