Kth Distinct String in an Array

Kth Distinct String in an Array - Problem

A distinct string is a string that is present only once in an array.

Given an array of strings arr, and an integer k, return the kth distinct string present in arr. If there are fewer than k distinct strings, return an empty string "".

Note: The strings are considered in the order in which they appear in the array.

Input & Output

Example 1 — Basic Case

$ Input: arr = ["d","b","c","b","c","a"], k = 2

› Output: "a"

💡 Note: The distinct strings are "d" and "a" (appear only once). The 2nd distinct string is "a".

Example 2 — Not Enough Distinct

$ Input: arr = ["aaa","aa","a"], k = 1

› Output: "aaa"

💡 Note: All strings are distinct (each appears once). The 1st distinct string is "aaa".

Example 3 — No kth Distinct

$ Input: arr = ["a","b","a"], k = 3

› Output: ""

💡 Note: Only "b" is distinct (appears once). Since k=3 but there's only 1 distinct string, return empty string.

Constraints

1 ≤ k ≤ arr.length ≤ 1000
1 ≤ arr[i].length ≤ 5
arr[i] consists of lowercase English letters

Visualization

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Asked in

a Amazon 25 M Microsoft 18

The key insight is to count the frequency of each string first, then find the kth string that appears exactly once. Best approach uses a hash map for O(n) time counting, then iterates through the original array to maintain order. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Count Each String

⏱️ Time: O(n²) Space: O(1)

For each string in the array, scan through the entire array to count how many times it appears. If it appears exactly once, increment our distinct counter until we reach the kth distinct string.

Frequency Count - Two Pass Hash Map

⏱️ Time: O(n) Space: O(n)

Use a hash map to count frequency of each string in the first pass. In the second pass, iterate through the array and find strings with frequency 1 until we reach the kth distinct string.

Brute Force - Count Each String — Algorithm Steps

Iterate through each string in the array
For each string, count its total occurrences
If count is 1, increment distinct counter
Return the kth distinct string or empty string

Visualization

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Step-by-Step Walkthrough

Check First String

Count 'd' appears 1 time - it's distinct #1

Check Second String

Count 'b' appears 2 times - not distinct

Continue Until k=2

Find 'a' appears 1 time - it's distinct #2

Code -

solution.c — C

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

char* solution(char** arr, int arrSize, int k) {
    int distinctCount = 0;
    static char result[1000];
    
    for (int i = 0; i < arrSize; i++) {
        // Count occurrences of arr[i]
        int count = 0;
        for (int j = 0; j < arrSize; j++) {
            if (strcmp(arr[j], arr[i]) == 0) {
                count++;
            }
        }
        
        // If it appears exactly once
        if (count == 1) {
            distinctCount++;
            if (distinctCount == k) {
                strcpy(result, arr[i]);
                return result;
            }
        }
    }
    
    strcpy(result, "");
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array (simplified)
    char** arr = malloc(1000 * sizeof(char*));
    int arrSize = 0;
    
    char* token = strtok(line, "[],\"\n");
    while (token != NULL) {
        if (strlen(token) > 0 && strcmp(token, " ") != 0) {
            arr[arrSize] = malloc(strlen(token) + 1);
            strcpy(arr[arrSize], token);
            arrSize++;
        }
        token = strtok(NULL, "[],\"\n");
    }
    
    int k;
    scanf("%d", &k);
    
    char* result = solution(arr, arrSize, k);
    printf("%s\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n strings, we scan the entire array of n strings to count occurrences

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Count Each String — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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