Longest Common Prefix of K Strings After Removal - Problem

You are given an array of strings words and an integer k. For each index i in the range [0, words.length - 1], find the length of the longest common prefix among any k strings (selected at distinct indices) from the remaining array after removing the i-th element.

Return an array answer, where answer[i] is the answer for the i-th element. If removing the i-th element leaves the array with fewer than k strings, answer[i] is 0.

Input & Output

Example 1 — Basic Case

$ Input: words = ["cat","car","card","care"], k = 2

› Output: [3,3,2,3]

💡 Note: Remove "cat": best 2-combination is ["car","card"] or ["car","care"] with prefix "car" (length 3). Remove "car": best is ["card","care"] with prefix "car" (length 3). Remove "card": best is ["cat","car"] with prefix "ca" (length 2). Remove "care": best is ["car","card"] with prefix "car" (length 3).

Example 2 — Insufficient Strings

$ Input: words = ["ab","bc"], k = 3

› Output: [0,0]

💡 Note: After removing any element, only 1 string remains, but k=3 is required, so answer is [0,0].

Example 3 — No Common Prefix

$ Input: words = ["abc","def","ghi"], k = 2

› Output: [0,0,0]

💡 Note: No two strings share a common prefix, so longest common prefix length is 0 for all removals.

Constraints

2 ≤ words.length ≤ 100
1 ≤ words[i].length ≤ 100
1 ≤ k ≤ words.length
words[i] consists of lowercase English letters only

Visualization

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Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is to use a trie data structure to efficiently find common prefixes without generating all k-combinations. The optimal approach builds a trie with count tracking, temporarily removes counts for excluded words, then performs DFS to find the deepest node with k available paths. Time: O(n × m × L), Space: O(n × L).

Common Approaches

✓ Brute Force - Check All Combinations

⏱️ Time: O(n × C(n-1,k) × m) Space: O(C(n-1,k))

For each index to remove, generate all possible combinations of k strings from remaining array, then find the longest common prefix among each combination and take the maximum.

Trie-Based Optimization

⏱️ Time: O(n × m × L) Space: O(n × L)

Build a trie from all strings, then for each removal, use DFS on the trie to find the deepest node that has at least k children paths, representing the longest common prefix among k strings.

Brute Force - Check All Combinations — Algorithm Steps

Step 1: For each index i, create array without words[i]
Step 2: Generate all combinations of k strings from remaining array
Step 3: Find longest common prefix for each combination
Step 4: Return maximum prefix length found

Visualization

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Step-by-Step Walkthrough

Remove Element

Remove words[i] from array

Generate Combinations

Create all possible k-combinations from remaining strings

Find Max Prefix

Calculate longest common prefix for each combination

Code -

solution.c — C

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

#define MAX_WORDS 100
#define MAX_LEN 100
#define MAX_COMBOS 10000

static char words[MAX_WORDS][MAX_LEN + 1];
static char remaining[MAX_WORDS][MAX_LEN + 1];
static int combinations[MAX_COMBOS][MAX_WORDS];
static int combCount;

void generateCombinations(int n, int k, int start, int current[], int pos) {
    if (pos == k) {
        for (int i = 0; i < k; i++) {
            combinations[combCount][i] = current[i];
        }
        combCount++;
        return;
    }
    
    for (int i = start; i < n; i++) {
        current[pos] = i;
        generateCombinations(n, k, i + 1, current, pos + 1);
    }
}

int getCommonPrefixLength(int indices[], int k) {
    if (k == 0) return 0;
    
    int minLen = strlen(remaining[indices[0]]);
    for (int i = 1; i < k; i++) {
        int len = strlen(remaining[indices[i]]);
        if (len < minLen) minLen = len;
    }
    
    int prefixLen = 0;
    for (int i = 0; i < minLen; i++) {
        char ch = remaining[indices[0]][i];
        int allMatch = 1;
        for (int j = 1; j < k; j++) {
            if (remaining[indices[j]][i] != ch) {
                allMatch = 0;
                break;
            }
        }
        if (allMatch) {
            prefixLen++;
        } else {
            break;
        }
    }
    
    return prefixLen;
}

int parseStringArray(char* str, char arr[][MAX_LEN + 1]) {
    int count = 0;
    if (strcmp(str, "[]") == 0) return 0;
    
    int inString = 0;
    int pos = 0;
    
    for (int i = 1; i < strlen(str) - 1; i++) {
        char c = str[i];
        if (c == '"') {
            if (inString) {
                arr[count][pos] = '\0';
                count++;
                pos = 0;
                inString = 0;
            } else {
                inString = 1;
            }
        } else if (inString) {
            arr[count][pos++] = c;
        }
    }
    
    return count;
}

int main() {
    char wordsStr[1000];
    int k;
    
    fgets(wordsStr, 1000, stdin);
    wordsStr[strlen(wordsStr) - 1] = '\0'; // remove newline
    scanf("%d", &k);
    
    int n = parseStringArray(wordsStr, words);
    
    printf("[");
    for (int i = 0; i < n; i++) {
        // Create remaining array without words[i]
        int remainingCount = 0;
        for (int j = 0; j < n; j++) {
            if (j != i) {
                strcpy(remaining[remainingCount], words[j]);
                remainingCount++;
            }
        }
        
        int result = 0;
        if (remainingCount >= k) {
            // Generate all combinations of k strings
            combCount = 0;
            int current[MAX_WORDS];
            generateCombinations(remainingCount, k, 0, current, 0);
            
            // Find maximum prefix length
            int maxPrefixLen = 0;
            for (int c = 0; c < combCount; c++) {
                int prefixLen = getCommonPrefixLength(combinations[c], k);
                if (prefixLen > maxPrefixLen) {
                    maxPrefixLen = prefixLen;
                }
            }
            result = maxPrefixLen;
        }
        
        printf("%d", result);
        if (i < n - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × C(n-1,k) × m)

For each of n removals, check C(n-1,k) combinations, each taking O(m) time to find common prefix

✓ Linear Growth

Space Complexity

O(C(n-1,k))

Space to store all k-combinations

⚡ Linearithmic Space

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Longest Common Prefix of K Strings After Removal - Problem

Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Combinations — Algorithm Steps

Visualization

Code -

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