Minimum Cost to Make All Characters Equal

Minimum Cost to Make All Characters Equal - Problem

You are given a 0-indexed binary string s of length n on which you can apply two types of operations:

Choose an index i and invert all characters from index 0 to index i (both inclusive), with a cost of i + 1
Choose an index i and invert all characters from index i to index n - 1 (both inclusive), with a cost of n - i

Return the minimum cost to make all characters of the string equal.

Invert a character means if its value is '0' it becomes '1' and vice-versa.

Input & Output

Example 1 — Basic Case

$ Input: s = "0011"

› Output: 2

💡 Note: There's one transition at position 1 (between s[1]='0' and s[2]='1'). To eliminate this transition, we can use either a left operation at index 1 (cost = 2) or right operation at index 2 (cost = 2). The minimum cost is 2.

Example 2 — All Same

$ Input: s = "1111"

› Output: 0

💡 Note: All characters are already equal, so no operations needed. Cost = 0.

Example 3 — Multiple Transitions

$ Input: s = "010"

› Output: 2

💡 Note: Two transitions: at positions 0 (0→1) and 1 (1→0). For each transition, we choose the cheaper operation. Total cost = min(1,2) + min(2,1) = 1 + 1 = 2.

Constraints

1 ≤ s.length ≤ 10⁵
s consists only of '0' and '1'

Visualization

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Asked in

G Google 25 f Meta 18 a Amazon 15

The key insight is that we only need to fix transition points where adjacent characters differ. For each transition, we choose the cheaper of two operations: flipping left (cost i+1) or flipping right (cost n-i). The optimal greedy approach runs in O(n) time with O(1) space.

Common Approaches

✓ Greedy Two-Pass Solution

⏱️ Time: O(n) Space: O(1)

Calculate the minimum cost to make all characters '0' and the minimum cost to make all characters '1', then return the smaller cost.

Try All Possible Operations

⏱️ Time: O(2^n) Space: O(n)

Generate all possible combinations of left and right operations and simulate each to find the minimum cost to make all characters equal.

Greedy Two-Pass Solution — Algorithm Steps

Calculate cost to make all characters '0'
Calculate cost to make all characters '1'
Return the minimum of both costs

Visualization

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Step-by-Step Walkthrough

Find Transitions

Identify positions where adjacent characters differ

Calculate Costs

For each transition, choose cheaper of left (i+1) or right (n-i-1) operation

Sum Minimum

Add up all minimum costs to get total

Code -

solution.c — C

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

int min(int a, int b) {
    return a < b ? a : b;
}

int solution(char* s) {
    int n = strlen(s);
    if (n == 1) return 0;
    
    int cost = 0;
    
    // Find all transition points and sum minimum costs
    for (int i = 0; i < n - 1; i++) {
        if (s[i] != s[i + 1]) {
            // At transition point, choose cheaper operation
            cost += min(i + 1, n - i - 1);
        }
    }
    
    return cost;
}

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(n)

Single pass through the string to find transitions and calculate costs

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for cost calculation

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Two-Pass Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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