Minimum Swaps to Arrange a Binary Grid - Practice Coding Problems

Minimum Swaps to Arrange a Binary Grid - Problem

Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them.

A grid is said to be valid if all the cells above the main diagonal are zeros.

Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.

The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).

Input & Output

Example 1 — Basic Grid Arrangement

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

› Output: 2

💡 Note: Row 2 [1,0,0,0] has 3 trailing zeros, perfect for position 0. We swap row 2 with row 1 (1 swap), then row 1 with row 0 (1 swap), giving us 2 total swaps.

Example 2 — Already Valid Grid

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

› Output: 0

💡 Note: The grid is already valid - all cells above the main diagonal are zeros. No swaps needed.

Example 3 — Impossible Case

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

› Output: -1

💡 Note: Row 0 needs ≥2 trailing zeros but [1,1,1] has 0. Row 2 [1,1,1] also has 0. No valid arrangement possible.

Constraints

n == grid.length
n == grid[i].length
1 ≤ n ≤ 200
grid[i][j] is 0 or 1

Visualization

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Understanding the Visualization

Input Grid

Binary grid with 1s above diagonal

Valid Constraint

All cells above main diagonal must be 0

Adjacent Swaps

Minimum swaps to achieve valid state

Key Takeaway

🎯 Key Insight: Count trailing zeros in each row - this determines the minimum position where each row can be validly placed

Asked in

G Google 15 a Amazon 12 M Microsoft 8 f Facebook 6

The key insight is counting trailing zeros in each row to determine valid positions. For a grid to be valid, row i must have at least n-1-i trailing zeros. The greedy approach works well: for each position, find the nearest suitable row and bubble it up using adjacent swaps. Best approach achieves Time: O(n²), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy - Count Trailing Zeros	O(n²)	O(n)	Count trailing zeros in each row and greedily place rows based on position requirements
Brute Force - Try All Permutations	O(n! × n²)	O(n²)	Generate all possible row arrangements and find minimum swaps

Greedy - Count Trailing Zeros — Algorithm Steps

Count trailing zeros for each row
For each position i (top to bottom), find nearest row with enough trailing zeros
Bubble the found row to position i using adjacent swaps

Visualization

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Step-by-Step Walkthrough

Count Trailing Zeros

Count zeros from right for each row

Position Requirements

Row i needs n-1-i trailing zeros

Bubble Suitable Rows

Move qualifying rows to correct positions

Code -

solution.c — C

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

int solution(int** grid, int gridSize, int* gridColSize) {
    int n = gridSize;
    
    // Count trailing zeros for each row
    int* trailingZeros = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        int count = 0;
        for (int j = gridColSize[i] - 1; j >= 0; j--) {
            if (grid[i][j] == 0) {
                count++;
            } else {
                break;
            }
        }
        trailingZeros[i] = count;
    }
    
    int swaps = 0;
    // For each position i, we need at least n-1-i trailing zeros
    for (int i = 0; i < n; i++) {
        int required = n - 1 - i;
        
        // Find the first row from position i onwards that has enough trailing zeros
        int found = -1;
        for (int j = i; j < n; j++) {
            if (trailingZeros[j] >= required) {
                found = j;
                break;
            }
        }
        
        if (found == -1) {
            free(trailingZeros);
            return -1; // No valid arrangement possible
        }
        
        // Bubble the found row to position i using adjacent swaps
        while (found > i) {
            int temp = trailingZeros[found];
            trailingZeros[found] = trailingZeros[found-1];
            trailingZeros[found-1] = temp;
            found--;
            swaps++;
        }
    }
    
    free(trailingZeros);
    return swaps;
}

int main() {
    // Simplified C implementation for 2D array input
    printf("3\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n positions, scan remaining rows and perform adjacent swaps

⚠ Quadratic Growth

Space Complexity

O(n)

Store trailing zero counts for each row

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy - Count Trailing Zeros — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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