Minimum Moves to Convert String - Practice Coding Problems

Minimum Moves to Convert String - Problem

String Greedy Easy

You are given a string s consisting of n characters which are either 'X' or 'O'.

A move is defined as selecting three consecutive characters of s and converting them to 'O'. Note that if a move is applied to the character 'O', it will stay the same.

Return the minimum number of moves required so that all the characters of s are converted to 'O'.

Input & Output

Example 1 — Basic Case

$ Input: s = "XXX"

› Output: 1

💡 Note: We can convert all three X's to O's with a single move: one move converts "XXX" to "OOO"

Example 2 — Mixed Characters

$ Input: s = "XXOX"

› Output: 2

💡 Note: First move at position 0 converts "XXOX" to "OOOX", second move at position 3 converts it to "OOOO"

Example 3 — Already Converted

$ Input: s = "OOOO"

› Output: 0

💡 Note: All characters are already 'O', so no moves are needed

Constraints

1 ≤ s.length ≤ 1000
s[i] is either 'X' or 'O'

Visualization

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Understanding the Visualization

Input

String with X's and O's

Process

Apply moves to convert 3 consecutive chars to O

Output

Minimum number of moves needed

Key Takeaway

🎯 Key Insight: Greedily convert X's from left to right - delaying never helps!

Asked in

M Microsoft 25 a Amazon 20

The key insight is that we can use a greedy approach: scan from left to right and make a move whenever we encounter an 'X'. This is optimal because delaying the move never helps. Best approach is greedy scanning with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy - Left to Right Scan	O(n)	O(1)	Scan left to right, make a move whenever we encounter 'X'
Brute Force - Try All Combinations	O(2ⁿ)	O(n)	Try all possible combinations of moves to find minimum

Greedy - Left to Right Scan — Algorithm Steps

Initialize move counter and position pointer
Scan string from left to right
When we find 'X', increment moves and skip 3 positions
Continue until end of string

Visualization

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Step-by-Step Walkthrough

Scan Position

Check current character

Make Move

If 'X' found, make move and skip 3 positions

Continue

Repeat until end of string

Code -

solution.c — C

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

int solution(char* s) {
    int moves = 0;
    int i = 0;
    int n = strlen(s);
    
    while (i < n) {
        if (s[i] == 'X') {
            moves++;
            i += 3; // Skip the next 2 characters as they're now 'O'
        } else {
            i++;
        }
    }
    
    return moves;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string, each character visited at most once

✓ Linear Growth

Space Complexity

O(1)

Only uses constant extra space for counters and pointers

✓ Linear Space

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Medium Frequency

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy - Left to Right Scan — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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