Optimal Placement of Dominant Entities — Problem Statement & Solution Guide

DesignHardN-Queens

TimeO(N! / 2^N) in practice, often expressed as O(N * 2^N) with bitmask pruning

SpaceO(N) for recursion stack and bitmask variables

Problem Description

Given an integer N, compute the total number of distinct ways to place N dominant entities on an N×N board such that no two entities share the same row, column, or diagonal. Each placement must satisfy the classic non‑attacking condition: for any two entities positioned at (r1, c1) and (r2, c2), the absolute differences |r1‑r2| and |c1‑c2| must never be equal. Return the count of all valid configurations. The board is empty at the start, and rotations or reflections of a configuration are considered different arrangements.

DSA Pattern Breakdown

"Optimal Placement of Dominant Entities"

hard

WHY DOES IT MATTER?

The N‑Queens counting problem exemplifies combinatorial search with exponential blow‑up, making it a perfect showcase for pruning techniques, bit manipulation, and symmetry exploitation—core skills for any senior engineer tackling large‑scale search or optimization tasks.

OPTIMIZATION CHALLENGE

The breakthrough is representing three independent constraints (column, major diagonal, minor diagonal) as integers and using bitwise operations to compute the intersection of free positions in O(1) per row, turning a factorial search into a dramatically pruned exponential one.

REAL-WORLD CONNECTION

Think of allocating exclusive resources (CPU cores, network ports, or database shards) across multiple services where each resource must not conflict on shared constraints; the bitmask approach mirrors how distributed schedulers quickly eliminate infeasible placements.

When coding under interview pressure, first write the simple backtrack, then immediately replace the three boolean arrays with bitmask integers; this single change often drops the runtime from seconds to milliseconds and demonstrates mastery of low‑level optimization.

COMPLEXITY AT A GLANCE

⏱ Time:O(N! / 2^N) in practice, often expressed as O(N * 2^N) with bitmask pruning

💾 Space:O(N) for recursion stack and bitmask variables

Core Theory — Why This Approach?

The problem is a classic counting variant of the N‑Queens puzzle. The board can be represented as a permutation of columns – one queen per row – and the diagonal constraints translate to two additional sets: the major diagonal (row‑col) and the minor diagonal (row+col). A naive back‑track that tries every column in every row leads to O(N!) possibilities, which explodes even for N≈14. The optimal paradigm uses depth‑first search with bitmask pruning: three bitsets track which columns, major diagonals, and minor diagonals are already occupied. At each recursive level we compute the set of safe positions with a single bitwise AND/NOT operation, then iterate over the set bits using the low‑bit extraction trick. This reduces the search space dramatically and enables counting solutions for N up to 15‑16 within milliseconds.

Interview Questions on This Problem

Q1How would you modify the bitmask solution to also return one valid board configuration for a given N?

Maintain an auxiliary array that records the column index chosen at each depth of recursion. When a full placement is found, copy this array into a result matrix (or list of strings). The bitmask logic stays unchanged; you just add bookkeeping for the chosen column before recursing.

Q2Explain why symmetry reduction (mirroring and rotation) can halve the runtime for even N, and how you would incorporate it.

The N‑Queens board is symmetric under 90°, 180°, and vertical/horizontal reflections. For even N, solutions can be paired with their mirror images, so you can count only placements where the queen in the first row is placed in the left half of columns and multiply the count appropriately, handling the central column separately for odd N. Implement this by limiting the initial column choices and applying a factor of 2 (or 4 for full symmetry) to the final count.

Q3What are the trade‑offs between using recursive backtracking with bitmasks versus an iterative stack‑based implementation for this problem?

Recursive backtracking is concise and leverages the call stack for state restoration, but deep recursion may hit language stack limits for large N. An iterative version uses an explicit stack to store row, column, and diagonal masks, avoiding recursion depth issues and often yielding slightly better performance due to reduced function‑call overhead, at the cost of more complex code.

Examples

Example 1

Input

Output

Explanation: With a single cell, the sole entity occupies it, forming one valid arrangement.

Example 2

Input

Output

Explanation: Two placements satisfy the constraints: 1. Entities at (0,1), (1,3), (2,0), (3,2). 2. Entities at (0,2), (1,0), (2,3), (3,1). Both sets use one entity per row and column, and no two share a diagonal.

Example 3

Input

Output

Explanation: Four distinct configurations exist for N=6. One example is: (0,1), (1,3), (2,5), (3,0), (4,2), (5,4). Repeating the verification for each configuration confirms that rows, columns, and both diagonal directions contain no conflicts, yielding a total of four solutions.

Constraints

1 <= N <= 14
The answer fits within a 64‑bit signed integer.

Optimal Approach & Strategy

Use depth‑first search with three bitmask integers to represent occupied columns and both diagonals, compute safe positions with bitwise operations, and recurse only on those bits, counting solutions as you reach the last row.

Brute Force Approach

Try every possible placement of N queens on the board (N! permutations) and check each for diagonal conflicts, which is exponential and infeasible for N > 10.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N! / 2^N) in practice, often expressed as O(N * 2^N) with bitmask pruning

function placeQueens(n) {
   const board = Array(n).fill(0).map(() => Array(n).fill('.'));
   const placeQueen = (row, col) => {
      board[row][col] = 'Q';
      for (let i = 0; i < n; i++) {
         if (i !== row && board[i][col] === 'Q') return false;
         if (i !== row && col - (i - row) >= 0 && board[i][col - (i - row)] === 'Q') return false;
         if (i !== row && col + (i - row) < n && board[i][col + (i - row)] === 'Q') return false;
      }
      return true;
   };
   const placeQueensHelper = (row) => {
      if (row === n) return true;
      for (let col = 0; col < n; col++) {
         if (placeQueen(row, col)) {
            if (placeQueensHelper(row + 1)) return true;
            board[row][col] = '.';
         }
      }
      return false;
   };
   placeQueensHelper(0);
   return board;
}

class Solution {
   public:
      vector<string> placeQueens(int n) {
         vector<string> board(n, string(n, '.'));
         return placeQueensHelper(0, board);
      }
   private:
      vector<string> placeQueensHelper(int row, vector<string>& board) {
         if (row == board.size()) {
            vector<string> result;
            for (int i = 0; i < board.size(); i++) {
               result.push_back(board[i]);
            }
            return result;
         }
         for (int col = 0; col < board[row].size(); col++) {
            if (placeQueen(row, col, board)) {
               if (placeQueensHelper(row + 1, board)) {
                  return placeQueensHelper(row + 1, board);
               }
               board[row][col] = '.';
            }
         }
         return {};
      }
   private:
      bool placeQueen(int row, int col, vector<string>& board) {
         for (int i = 0; i < board.size(); i++) {
            if (i != row && board[i][col] == 'Q') {
               return false;
            }
            if (i != row && col - (i - row) >= 0 && board[i][col - (i - row)] == 'Q') {
               return false;
            }
            if (i != row && col + (i - row) < board[row].size() && board[i][col + (i - row)] == 'Q') {
               return false;
            }
         }
         return true;
      }
}

public class Solution {
   public String[] placeQueens(int n) {
      char[][] board = new char[n][n];
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < n; j++) {
            board[i][j] = '.';
         }
      }
      return placeQueensHelper(0, board);
   }
   private String[] placeQueensHelper(int row, char[][] board) {
      if (row == board.length) {
         String[] result = new String[board.length];
         for (int i = 0; i < board.length; i++) {
            StringBuilder sb = new StringBuilder();
            for (int j = 0; j < board[i].length; j++) {
               sb.append(board[i][j]);
            }
            result[i] = sb.toString();
         }
         return result;
      }
      for (int col = 0; col < board[row].length; col++) {
         if (placeQueen(row, col, board)) {
            if (placeQueensHelper(row + 1, board)) {
               return placeQueensHelper(row + 1, board);
            }
            board[row][col] = '.';
         }
      }
      return null;
   }
   private boolean placeQueen(int row, int col, char[][] board) {
      for (int i = 0; i < board.length; i++) {
         if (i != row && board[i][col] == 'Q') {
            return false;
         }
         if (i != row && col - (i - row) >= 0 && board[i][col - (i - row)] == 'Q') {
            return false;
         }
         if (i != row && col + (i - row) < board[row].length && board[i][col + (i - row)] == 'Q') {
            return false;
         }
      }
      return true;
   }
}

def place_queens(n):
   board = [['.' for _ in range(n)] for _ in range(n)]
   def place_queen(row, col):
      board[row][col] = 'Q'
      for i in range(n):
         if i != row and board[i][col] == 'Q':
            return False
         if i != row and col - (i - row) >= 0 and board[i][col - (i - row)] == 'Q':
            return False
         if i != row and col + (i - row) < n and board[i][col + (i - row)] == 'Q':
            return False
      return True
   def place_queens_helper(row):
      if row == n:
         return True
      for col in range(n):
         if place_queen(row, col):
            if place_queens_helper(row + 1):
               return True
            board[row][col] = '.'
      return False
   place_queens_helper(0)
   return board

function placeQueens(n) {
   const board = Array(n).fill(0).map(() => Array(n).fill('.'));
   const placeQueen = (row, col) => {
      board[row][col] = 'Q';
      for (let i = 0; i < n; i++) {
         if (i !== row && board[i][col] === 'Q') return false;
         if (i !== row && col - (i - row) >= 0 && board[i][col - (i - row)] === 'Q') return false;
         if (i !== row && col + (i - row) < n && board[i][col + (i - row)] === 'Q') return false;
      }
      return true;
   };
   const placeQueensHelper = (row) => {
      if (row === n) return true;
      for (let col = 0; col < n; col++) {
         if (placeQueen(row, col)) {
            if (placeQueensHelper(row + 1)) return true;
            board[row][col] = '.';
         }
      }
      return false;
   };
   placeQueensHelper(0);
   return board;
}

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Optimal Placement of Dominant Entities Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Design

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Optimal Placement of Dominant Entities — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Solve in Interative Editor