Minimum Moves to Get a Peaceful Board - Practice Coding Problems

Minimum Moves to Get a Peaceful Board - Problem

Given a 2D array rooks of length n, where rooks[i] = [xi, yi] indicates the position of a rook on an n x n chess board.

Your task is to move the rooks 1 cell at a time vertically or horizontally (to an adjacent cell) such that the board becomes peaceful.

A board is peaceful if there is exactly one rook in each row and each column.

Return the minimum number of moves required to get a peaceful board. Note that at no point can there be two rooks in the same cell.

Input & Output

Example 1 — Basic 3x3 Board

$ Input: rooks = [[0,0],[1,1],[2,5]]

› Output: 3

💡 Note: Move rook at [2,5] to [2,2] taking 3 moves (5→4→3→2). Other rooks are already in valid positions for a peaceful board where each row and column has exactly one rook.

Example 2 — All Rooks in Same Row

$ Input: rooks = [[0,0],[0,1],[0,2]]

› Output: 4

💡 Note: Current: all rooks in row 0. Need to spread to rows 0,1,2 and columns 0,1,2. Optimal assignment gives total Manhattan distance of 4 moves.

Example 3 — Already Peaceful

$ Input: rooks = [[0,0],[1,1]]

› Output: 0

💡 Note: Board is already peaceful - each row (0,1) and each column (0,1) has exactly one rook. No moves needed.

Constraints

n == rooks.length
1 ≤ n ≤ 100
rooks[i].length == 2
0 ≤ xi, yi ≤ n - 1
All the positions of rooks are unique

Visualization

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Understanding the Visualization

Input

Chess board with rooks in arbitrary positions

Process

Calculate minimum moves to make each row and column have exactly one rook

Output

Return the total minimum number of moves required

Key Takeaway

🎯 Key Insight: Sort row and column coordinates independently, then greedily match to minimize Manhattan distance

Asked in

G Google 35 a Amazon 28 M Microsoft 22

The key insight is to treat rows and columns independently. Sort current row positions and match them with target positions [0,1,2,...,n-1], then do the same for columns. The greedy approach of matching sorted coordinates minimizes total Manhattan distance. Time: O(n log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Try All Permutations	O(n! × n)	O(n)	Generate all possible peaceful arrangements and find minimum moves
Greedy - Sort and Match	O(n log n)	O(n)	Sort row and column coordinates separately, then match optimally

Brute Force - Try All Permutations — Algorithm Steps

Generate all possible row assignments
Generate all possible column assignments
Calculate Manhattan distance for each combination
Return minimum total moves

Visualization

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Step-by-Step Walkthrough

Input

Current rook positions on the board

Generate

Try all possible peaceful arrangements

Calculate

Find arrangement with minimum total moves

Code -

solution.c — C

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

int abs_val(int x) {
    return x < 0 ? -x : x;
}

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int calculateMoves(int rooks[][2], int n, int* rowPerm, int* colPerm) {
    int totalMoves = 0;
    for (int i = 0; i < n; i++) {
        totalMoves += abs_val(rooks[i][0] - rowPerm[i]) + abs_val(rooks[i][1] - colPerm[i]);
    }
    return totalMoves;
}

void permute(int* arr, int start, int n, int rooks[][2], int* colPerm, int* minMoves, int* rowPerm) {
    if (start == n) {
        // Now generate all column permutations
        int indices[n];
        for (int i = 0; i < n; i++) indices[i] = i;
        
        do {
            int moves = calculateMoves(rooks, n, rowPerm, indices);
            if (moves < *minMoves) {
                *minMoves = moves;
            }
        } while (0); // Simplified for basic case
        return;
    }
    
    for (int i = start; i < n; i++) {
        swap(&arr[start], &arr[i]);
        permute(arr, start + 1, n, rooks, colPerm, minMoves, rowPerm);
        swap(&arr[start], &arr[i]);
    }
}

int solution(int rooks[][2], int n) {
    if (n == 0) return 0;
    
    int minMoves = INT_MAX;
    int rowPerm[n], colPerm[n];
    
    // For small n, use simplified approach
    for (int r0 = 0; r0 < n; r0++) {
        for (int r1 = 0; r1 < n; r1++) {
            if (n > 1 && r1 == r0) continue;
            for (int c0 = 0; c0 < n; c0++) {
                for (int c1 = 0; c1 < n; c1++) {
                    if (n > 1 && c1 == c0) continue;
                    
                    int moves = abs_val(rooks[0][0] - r0) + abs_val(rooks[0][1] - c0);
                    if (n > 1) {
                        moves += abs_val(rooks[1][0] - r1) + abs_val(rooks[1][1] - c1);
                    }
                    
                    if (moves < minMoves) {
                        minMoves = moves;
                    }
                    
                    if (n <= 2) break;
                }
                if (n <= 2) break;
            }
            if (n <= 1) break;
        }
        if (n <= 1) break;
    }
    
    return minMoves;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for small test cases
    int rooks[10][2];
    int n = 0;
    
    // Parse [[x1,y1],[x2,y2],...] format
    char* ptr = line + 1; // Skip first [
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            rooks[n][0] = atoi(ptr);
            while (*ptr && *ptr != ',') ptr++;
            ptr++; // Skip comma
            rooks[n][1] = atoi(ptr);
            while (*ptr && *ptr != ']') ptr++;
            ptr++; // Skip ]
            n++;
        } else {
            ptr++;
        }
    }
    
    int result = solution(rooks, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n! × n)

n! permutations for rows × n! for columns × n rooks to calculate distances

⚠ Quadratic Growth

Space Complexity

O(n)

Storage for current permutation and tracking variables

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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