Maximum 69 Number - Practice Coding Problems

Maximum 69 Number - Problem

Math Greedy Easy

You are given a positive integer num consisting only of digits 6 and 9.

Return the maximum number you can get by changing at most one digit (6 becomes 9, and 9 becomes 6).

Input & Output

Example 1 — Basic Case with 6

$ Input: num = 9669

› Output: 9969

💡 Note: Change the first 6 to 9: 9669 → 9969. This gives maximum value since 6→9 increases the hundreds place.

Example 2 — All 9s

$ Input: num = 9996

› Output: 9999

💡 Note: Change the last 6 to 9: 9996 → 9999. Only one 6 present, changing it maximizes the number.

Example 3 — No 6s present

$ Input: num = 9999

› Output: 9996

💡 Note: No 6s to change to 9s. Change the rightmost 9 to 6: 9999 → 9996 for maximum with constraint.

Constraints

1 ≤ num ≤ 10⁴
num consists only of 6 and 9 digits

Visualization

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Understanding the Visualization

Input

Given number with only 6s and 9s

Strategy

Find leftmost 6 and change to 9

Output

Maximum possible number

Key Takeaway

🎯 Key Insight: The leftmost digit has the highest place value, so changing the first 6 to 9 maximizes the result

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that to maximize the number, we should change the leftmost 6 to 9 (highest place value impact). If no 6 exists, change the rightmost 9 to 6. The greedy approach achieves optimal O(n) time complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Try All Possibilities	O(n²)	O(n)	Generate all possible numbers by changing each digit and find maximum
Greedy First-6-to-9	O(n)	O(n)	Find the first 6 from left and change it to 9 for maximum value

Try All Possibilities — Algorithm Steps

Convert number to string to access individual digits
For each position, create new number with digit flipped
Convert each possibility back to integer
Return the maximum value found

Visualization

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Step-by-Step Walkthrough

Original

Start with given number

Try Each

Change each digit and compare

Maximum

Return the largest result

Code -

solution.c — C

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

int solution(int num) {
    char s[20];
    sprintf(s, "%d", num);
    int maxNum = num;
    int len = strlen(s);
    
    // Try changing each digit
    for (int i = 0; i < len; i++) {
        char temp[20];
        strcpy(temp, s);
        
        if (temp[i] == '6') {
            temp[i] = '9';
        } else { // temp[i] == '9'
            temp[i] = '6';
        }
        
        int newNum = atoi(temp);
        if (newNum > maxNum) {
            maxNum = newNum;
        }
    }
    
    return maxNum;
}

int main() {
    int num;
    scanf("%d", &num);
    
    int result = solution(num);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

O(n) loop to try each position × O(n) string operations for each conversion

⚠ Quadratic Growth

Space Complexity

O(n)

Multiple string copies created during conversions

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Try All Possibilities — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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