Find Maximum Number of String Pairs

Find Maximum Number of String Pairs - Problem

You are given a 0-indexed array words consisting of distinct strings.

The string words[i] can be paired with the string words[j] if:

The string words[i] is equal to the reversed string of words[j].
0 <= i < j < words.length.

Return the maximum number of pairs that can be formed from the array words.

Note that each string can belong in at most one pair.

Input & Output

Example 1 — Basic Case

$ Input: words = ["lc","cl","gg"]

› Output: 1

💡 Note: The string "lc" can be paired with "cl" since "lc" equals the reverse of "cl". We have 1 pair.

Example 2 — Multiple Pairs

$ Input: words = ["ab","ba","cc"]

› Output: 1

💡 Note: "ab" can pair with "ba" (reverse match). "cc" is its own reverse but needs another "cc" to pair, so it remains unpaired. Total: 1 pair.

Example 3 — No Pairs

$ Input: words = ["aa","ab"]

› Output: 0

💡 Note: "aa" is palindrome but needs another "aa" to pair. "ab" would need "ba" but it doesn't exist. No pairs possible.

Constraints

1 ≤ words.length ≤ 50
words[i].length == 2
words consists of distinct strings
words[i] contains only lowercase English letters

Visualization

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Asked in

A Amazon 15 G Google 12

The key insight is that each string can only pair with its exact reverse. The optimal approach uses a hash map to store seen strings and check for reverse matches in O(1) time. Best approach: Hash Map with Time: O(n×m), Space: O(n×m).

Common Approaches

✓ Brute Force

⏱️ Time: O(n² × m) Space: O(1)

Compare each string with every other string that comes after it. For each pair (i, j) where i < j, check if words[i] equals the reverse of words[j]. Count valid pairs and ensure each string is used at most once.

Hash Map Optimization

⏱️ Time: O(n × m) Space: O(n × m)

Iterate through the array once, for each string check if its reverse already exists in a hash map. If found, increment pair count and remove the reverse. Otherwise, add current string to the map.

Brute Force — Algorithm Steps

Iterate through each string i
For each i, check all strings j where j > i
If words[i] equals reverse of words[j], count as a pair
Mark both strings as used

Visualization

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Step-by-Step Walkthrough

Compare Each Pair

Check if words[i] equals reverse of words[j]

Mark Used

When pair found, mark both strings as used

Count Pairs

Count total valid pairs found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

void reverse_string(char* str, char* result) {
    int len = strlen(str);
    for (int i = 0; i < len; i++) {
        result[i] = str[len - 1 - i];
    }
    result[len] = '\0';
}

int solution(char words[][100], int n) {
    bool used[n];
    for (int i = 0; i < n; i++) used[i] = false;
    
    int pairs = 0;
    char reversed[100];
    
    for (int i = 0; i < n; i++) {
        if (used[i]) continue;
        for (int j = i + 1; j < n; j++) {
            if (used[j]) continue;
            reverse_string(words[j], reversed);
            if (strcmp(words[i], reversed) == 0) {
                pairs++;
                used[i] = true;
                used[j] = true;
                break;
            }
        }
    }
    
    return pairs;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char words[50][100];
    int n = 0;
    
    char* token = strtok(line, "[],\"\n");
    while (token != NULL && strlen(token) > 0) {
        if (token[0] != ' ' && token[0] != '\0') {
            strcpy(words[n], token);
            n++;
        }
        token = strtok(NULL, "[],\"\n");
    }
    
    int result = solution(words, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

Nested loops check all pairs (n²) and string reversal takes O(m) time

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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