Minimize String Length

Minimize String Length - Problem

Given a string s, you have two types of operation:

Choose an index i in the string, and let c be the character in position i. Delete the closest occurrence of c to the left of i (if exists).
Choose an index i in the string, and let c be the character in position i. Delete the closest occurrence of c to the right of i (if exists).

Your task is to minimize the length of s by performing the above operations zero or more times.

Return an integer denoting the length of the minimized string.

Input & Output

Example 1 — Basic Case

$ Input: s = "ababc"

› Output: 3

💡 Note: We can remove one 'a' and one 'b' using the operations, leaving "abc" with length 3. Since all remaining characters are unique, no further reduction is possible.

Example 2 — All Same Characters

$ Input: s = "aaa"

› Output: 1

💡 Note: We can remove the duplicates by choosing any 'a' and removing its neighbors. Eventually only one 'a' remains, giving length 1.

Example 3 — Already Unique

$ Input: s = "abc"

› Output: 3

💡 Note: All characters are already unique, so no operations are needed. The length remains 3.

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters

Visualization

Tap to expand

Asked in

M Microsoft 15 a Amazon 12

The key insight is that no matter how we perform the deletion operations, we can only keep one instance of each unique character. The optimal approach uses a set to count unique characters in O(n) time and O(1) space for lowercase letters.

Common Approaches

✓ Simulation Approach

⏱️ Time: O(n³) Space: O(n)

For each character in the string, try to find and remove its closest duplicate (left or right). Repeat this process until no more characters can be removed.

Unique Character Count

⏱️ Time: O(n) Space: O(k)

The key insight is that no matter how we remove duplicates, we can only keep one instance of each unique character. So the answer is simply the count of distinct characters.

Simulation Approach — Algorithm Steps

Scan string for characters with duplicates
Remove closest duplicate for each character
Repeat until no changes occur
Return final string length

Visualization

Tap to expand

Step-by-Step Walkthrough

Initial String

Start with original string

Find Duplicates

For each character, find closest duplicate

Remove & Repeat

Remove duplicates and repeat until no changes

Code -

solution.c — C

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

int solution(char* s) {
    int len = strlen(s);
    char* chars = malloc((len + 1) * sizeof(char));
    
    while (1) {
        int originalLen = strlen(s);
        strcpy(chars, s);
        int currentLen = originalLen;
        int i = 0;
        
        while (i < currentLen) {
            char c = chars[i];
            int found = 0;
            
            // Try to remove closest left duplicate
            for (int j = i - 1; j >= 0; j--) {
                if (chars[j] == c) {
                    for (int k = j; k < currentLen - 1; k++) {
                        chars[k] = chars[k + 1];
                    }
                    chars[--currentLen] = '\0';
                    i--;
                    found = 1;
                    break;
                }
            }
            
            if (!found) {
                // Try to remove closest right duplicate
                for (int j = i + 1; j < currentLen; j++) {
                    if (chars[j] == c) {
                        for (int k = j; k < currentLen - 1; k++) {
                            chars[k] = chars[k + 1];
                        }
                        chars[--currentLen] = '\0';
                        break;
                    }
                }
            }
            i++;
        }
        
        if (currentLen == originalLen) break;
        strcpy(s, chars);
    }
    
    free(chars);
    return strlen(s);
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n characters, we scan the string (O(n)) and potentially rebuild it (O(n)), repeated until convergence

⚠ Quadratic Growth

Space Complexity

O(n)

Creating new strings during the simulation process

⚡ Linearithmic Space

23.5K Views

Medium Frequency

~8 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simulation Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler