Find the Maximum Achievable Number

Find the Maximum Achievable Number - Problem

Math Easy

Given two integers, num and t. A number x is achievable if it can become equal to num after applying the following operation at most t times:

Operation: Increase or decrease x by 1, and simultaneously increase or decrease num by 1.

Return the maximum possible value of x.

Input & Output

Example 1 — Basic Case

$ Input: num = 4, t = 1

› Output: 6

💡 Note: Start with x = 6. Operation: x decreases to 5, num increases to 5. Now x = num = 5. Maximum achievable x was 6.

Example 2 — Multiple Operations

$ Input: num = 3, t = 2

› Output: 7

💡 Note: Start with x = 7. After 2 operations moving optimally: x can decrease to 5, num can increase to 5. Maximum achievable x was 7.

Example 3 — Zero Operations

$ Input: num = 5, t = 0

› Output: 5

💡 Note: No operations allowed, so x must equal num = 5 from the start. Maximum x is 5.

Constraints

1 ≤ num ≤ 1000
1 ≤ t ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that both x and num can move in opposite directions simultaneously. To maximize x, we start with x as large as possible and use all t operations to make x and num meet. The optimal strategy gives us the formula: maximum x = num + 2*t. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force Simulation

⏱️ Time: O(t²) Space: O(1)

For each possible starting value of x, simulate up to t operations to see if x can equal num. Track the maximum achievable x.

Mathematical Formula

⏱️ Time: O(1) Space: O(1)

Recognize that to maximize x, both x and num should move toward each other optimally. The maximum x is achieved when x increases by t and num decreases by t simultaneously.

Brute Force Simulation — Algorithm Steps

Try different starting values for x
For each x, simulate operations to make x equal num
Track maximum valid x found

Visualization

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Step-by-Step Walkthrough

Try x=1

Simulate operations from x=1

Try x=2

Simulate operations from x=2

Find Max

Return maximum valid x

Code -

solution.c — C

#include <stdio.h>

int solution(int num, int t) {
    // The maximum achievable x is num + 2*t
    // This is because we can start with x = num + 2*t
    // Then decrease x by t and increase num by t
    // They will meet at num + t
    return num + 2 * t;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(t²)

For each of O(t) possible x values, we simulate O(t) operations

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track current state

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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