Minimum Operations to Make a Special Number

Minimum Operations to Make a Special Number - Problem

You are given a 0-indexed string num representing a non-negative integer.

In one operation, you can pick any digit of num and delete it. Note that if you delete all the digits of num, num becomes 0.

Return the minimum number of operations required to make num special.

An integer x is considered special if it is divisible by 25.

Input & Output

Example 1 — Basic Pattern

$ Input: num = "2245047"

› Output: 2

💡 Note: We can form "2250" by deleting the digits '4' and '7'. Since 2250 % 25 = 0, we need 2 operations.

Example 2 — Single Zero

$ Input: num = "2908305"

› Output: 3

💡 Note: We can form "2900" by deleting '8', '3', '5'. Since 2900 % 25 = 0, we need 3 operations.

Example 3 — Already Special

$ Input: num = "10"

› Output: 1

💡 Note: 10 is not divisible by 25. We delete '1' to get '0', and 0 % 25 = 0, so we need 1 operation.

Constraints

1 ≤ num.length ≤ 10⁵
num consists of only digits
num does not contain any leading zeros

Visualization

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Asked in

G Google 23 a Amazon 18 M Microsoft 15

The key insight is that numbers divisible by 25 must end in 00, 25, 50, or 75. Best approach is greedy pattern matching - scan from right to find these patterns with minimum deletions. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Combinations

⏱️ Time: O(2ⁿ) Space: O(n)

For each possible subsequence of digits, check if the resulting number is divisible by 25. Track the minimum deletions needed.

Greedy Pattern Matching

⏱️ Time: O(n) Space: O(1)

Numbers divisible by 25 must end in 00, 25, 50, or 75. Scan from right to find these patterns with minimum deletions.

Brute Force - Try All Combinations — Algorithm Steps

Generate all possible subsequences
Convert each to number
Check if divisible by 25
Return minimum deletions

Visualization

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Step-by-Step Walkthrough

Generate Masks

Create all 2^n binary masks

Build Subsequences

For each mask, select corresponding digits

Check Divisibility

Test if result is divisible by 25

Code -

solution.c — C

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

int popCount(int x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

int solution(char* num) {
    int n = strlen(num);
    int minOps = n; // worst case: delete all digits to get 0
    
    // Try all possible subsequences (2^n combinations)
    for (int mask = 0; mask < (1 << n); mask++) {
        char subsequence[100] = {0};
        int idx = 0;
        
        for (int i = 0; i < n; i++) {
            if (mask & (1 << i)) {
                subsequence[idx++] = num[i];
            }
        }
        
        if (idx == 0) {
            strcpy(subsequence, "0");
        }
        
        // Remove leading zeros except for "0"
        if (strlen(subsequence) > 1) {
            int start = 0;
            while (subsequence[start] == '0' && start < strlen(subsequence)) {
                start++;
            }
            if (start == strlen(subsequence)) {
                strcpy(subsequence, "0");
            } else {
                memmove(subsequence, subsequence + start, strlen(subsequence) - start + 1);
            }
        }
        
        // Check if divisible by 25
        long long number = atoll(subsequence);
        if (number % 25 == 0) {
            int deletions = n - popCount(mask);
            if (deletions < minOps) {
                minOps = deletions;
            }
        }
    }
    
    return minOps;
}

int main() {
    char num[100];
    fgets(num, sizeof(num), stdin);
    num[strcspn(num, "\n")] = 0;
    
    int result = solution(num);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ)

Generate 2^n subsequences and check each one

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth and string storage

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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