Given a matrix A[][] of size N * N, where each matrix element is either -1 or 1. One can move from (X, Y) to either (X + 1, Y) or (X, Y + 1) but not out of the matrix. The task is to check whether a path exists from the top-left to the bottom-right corner such that the sum of all elements in the path is X.
Examples:
Input: arr[][] = {{1, 1, 1, 1}, {-1, 1, -1, -1}, {-1, 1, -1, 1}, {-1, -1, 1, 1}}, X = 3
Output: Yes
Explanation:
One path will be :
Cell 1: (0, 0), Sum = 1 now moving down to (1, 0)
Cell 2: (1, 0), Sum = 1 – 1 now moving right to (1, 1)
Cell 3: (1, 1), Sum = 1 – 1 + 1 now moving down to (2, 1)
Cell 4: (2, 1), Sum = 1 – 1 + 1 + 1 now moving down to (3, 1)
Cell 5: (3, 1), Sum = 1 – 1 + 1 + 1 – 1 now moving right to (3, 2)
Cell 6: (3, 2), Sum = 1 – 1 + 1 + 1 – 1 + 1 now moving right to (3, 3)
Cell 7: (3, 3), Sum = 1 – 1 + 1 + 1 – 1 + 1 + 1
The total sum is 3 which is equal to X Hence Print “Yes”. For a detailed explanation check the Efficient approach.Input: arr[][] = {{1, 1, 1}, {1, 1, 1}, {1, 1, 1}}, X = 4
Output: No
Naive approach: The basic way to solve the problem is as follows:
Visit all paths and track their sum by using recursive brute force in exponential time.
Time Complexity: O(2N * N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following idea
Let’s say minPathSum and maxPathSum are the maximum and minimum path sum from (0, 0) to (N-1, N-1) which can be calculated by Dynamic programming.
- dp[i][j] is minimum/maximum path sum at cell (i, j)
- recurrence relation : dp[i][j] = min or max of (dp[i – 1][j], dp[i][j – 1]) + A[i][j]
In this problem most important thing is greedy observation below.
- if number of cells from top-left to bottom-right is odd then all path sums of odd values can be generated in range [minPathSum, maxPathSum]
- if number of cells from top-left to bottom-right are even then all path sum of even values can be generated in range [minPathSum, maxPathSum]
Follow the steps below to solve the problem:
- Calculating minPathSum and maxPathSum using dynamic programming.
- Calculating the number of cells from (1, 1) to (N, N) with the formula numberOfCells = 2 * N – 1.
- if X has the same parity(fact of being odd or even) as numberOfCells and X is in between range [minPathSum, maxPathSum] print “Yes” else print “No”.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to check whether paths// exists from (1, 1) to (N, N) such// that its sum is Xvoid isReachable(vector<vector<int> > A, int N, int X){ // DP table for finding maximum and // minimum path sum vector<vector<vector<int> > > dp( N, vector<vector<int> >(N, vector<int>(2, 0))); // Base Case dp[0][0][0] = A[0][0]; dp[0][0][1] = A[0][0]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // Base Case if (i == 0 and j == 0) continue; // When coming from left is not // possible if (i == 0) { dp[i][j][0] = dp[i][j - 1][0] + A[i][j - 1]; dp[i][j][1] = dp[i][j - 1][1] + A[i][j - 1]; } // When coming from Left is not // possible else if (j == 0) { dp[i][j][0] = dp[i - 1][j][0] + A[i - 1][j]; dp[i][j][1] = dp[i - 1][j][1] + A[i - 1][j]; } // When Both coming from Left and Up // possible else { dp[i][j][0] = min(dp[i - 1][j][0] + A[i - 1][j], dp[i][j - 1][0] + A[i][j - 1]); dp[i][j][1] = max(dp[i - 1][j][1] + A[i - 1][j], dp[i][j - 1][1] + A[i][j - 1]); } } } // Minimum Path sum from (1, 1) to (N, N) int minimumPathSum = dp[N - 1][N - 1][0]; // Maximum Path sum from (1, 1) to (N, N) int maximumPathSum = dp[N - 1][N - 1][1]; // Number of cells in path from (1, 1) to (N, N) int numberOfCells = 2 * N - 1; // Parity of both numberOfCells from (1, 1) // to (N, N) and X should be same and X // should be in between range from // minmumPathSum and maximumPathSum if (numberOfCells % 2 == X % 2 && minimumPathSum <= X && maximumPathSum >= X) cout << "Yes" << endl; else cout << "No" << endl;}// Driver Codeint main(){ // Input 1 vector<vector<int> > A{ { 1, 1, 1, 1 }, { -1, 1, -1, -1 }, { -1, 1, -1, 1 }, { -1, -1, 1, 1 } }; int N = A.size(); int X = 3; // Function Call isReachable(A, N, X); // Input 2 vector<vector<int> > A1{ { 1, 1, 1 }, { 1, 1, 1 }, { 1, 1, 1 } }; int N1 = A1.size(); int X1 = 4; // Function Call isReachable(A1, N1, X1); return 0;} |
Java
// Java code to implement the approachclass GFG { // Function to check whether paths exists from (1, 1) to // (N, N) such that its sum is X public static void isReachable(int[][] A, int N, int X) { // DP table for finding maximum and minimum path sum int[][][] dp = new int[N][N][2]; // Base Case dp[0][0][0] = A[0][0]; dp[0][0][1] = A[0][0]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // Base Case if (i == 0 && j == 0) { continue; } // When coming from left is not possible if (i == 0) { dp[i][j][0] = dp[i][j - 1][0] + A[i][j - 1]; dp[i][j][1] = dp[i][j - 1][1] + A[i][j - 1]; } // When coming from Left is not possible else if (j == 0) { dp[i][j][0] = dp[i - 1][j][0] + A[i - 1][j]; dp[i][j][1] = dp[i - 1][j][1] + A[i - 1][j]; } // When Both coming from Left and Up possible else { dp[i][j][0] = Math.min(dp[i - 1][j][0] + A[i - 1][j], dp[i][j - 1][0] + A[i][j - 1]); dp[i][j][1] = Math.max(dp[i - 1][j][1] + A[i - 1][j], dp[i][j - 1][1] + A[i][j - 1]); } } } // Minimum Path sum from (1, 1) to (N, N) int minimumPathSum = dp[N - 1][N - 1][0]; // Maximum Path sum from (1, 1) to (N, N) int maximumPathSum = dp[N - 1][N - 1][1]; // Number of cells in path from (1, 1) to (N, N) int numberOfCells = 2 * N - 1; // Parity of both numberOfCells from (1, 1) // to (N, N) and X should be same and X // should be in between range from // minmumPathSum and maximumPathSum if (numberOfCells % 2 == X % 2 && minimumPathSum <= X && maximumPathSum >= X) { System.out.println("Yes"); } else { System.out.println("No"); } } // Driver Code public static void main(String[] args) { int[][] A = { { 1, 1, 1, 1 }, { -1, 1, -1, -1 }, { -1, 1, -1, 1 }, { -1, -1, 1, 1 } }; int N = 4; int X = 3; // Function Call isReachable(A, N, X); // Input 2 int[][] A1 = { { 1, 1, 1, 1 }, { -1, 1, -1, -1 }, { -1, 1, -1, 1 }, { -1, -1, 1, 1 } }; int N1 = 3; int X1 = 4; // Function Call isReachable(A1, N1, X1); }} |
Python3
# Python code to implement the approachimport math# Function to check whether paths exists from # (1, 1) to (N, N) such that its sum is xdef is_reachable(A, N, X): # DP table for finding maximum and minimum path sum dp = [[[0 for i in range(2)] for j in range(N)] for k in range(N)] # Base Case dp[0][0][0] = A[0][0] dp[0][0][1] = A[0][0] for i in range(N): for j in range(N): # Base Case if i == 0 and j == 0: continue # When coming from left is not possible elif i == 0: dp[i][j][0] = dp[i][j-1][0] + A[i][j-1] dp[i][j][1] = dp[i][j-1][1] + A[i][j-1] # When coming from Left is not possible elif j == 0: dp[i][j][0] = dp[i-1][j][0] + A[i-1][j] dp[i][j][1] = dp[i-1][j][1] + A[i-1][j] # When Both coming from Left and Up possible else: dp[i][j][0] = min(dp[i-1][j][0] + A[i-1][j], dp[i][j-1][0] + A[i][j-1]) dp[i][j][1] = max(dp[i-1][j][1] + A[i-1][j], dp[i][j-1][1] + A[i][j-1]) # Minimum Path sum from (1, 1) to (N, N) minimum_path_sum = dp[N-1][N-1][0] # Maximum Path sum from (1, 1) to (N, N) maximum_path_sum = dp[N-1][N-1][1] # Number of cells in path from (1, 1) to (N, N) number_of_cells = 2 * N - 1 # Parity of both numberOfCells from (1, 1) # to (N, N) and X should be same and X # should be in between range from # minmumPathSum and maximumPathSum if number_of_cells % 2 == X % 2 and minimum_path_sum <= X and maximum_path_sum >= X: print("Yes") else: print("No")# Driver CodeA = [[1, 1, 1, 1], [-1, 1, -1, -1], [-1, 1, -1, 1], [-1, -1, 1, 1]]N = 4X = 3# Function Callis_reachable(A, N, X)# Input 2A1 = [[1, 1, 1, 1], [-1, 1, -1, -1], [-1, 1, -1, 1], [-1, -1, 1, 1]]N1 = 3X1 = 4# Function Callis_reachable(A1, N1, X1)# This code is contributed by lokeshmvs21. |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG { // Function to check whether paths // exists from (1, 1) to (N, N) such // that its sum is X static void isReachable(List<List<int> > A, int N, int X) { // DP table for finding maximum and // minimum path sum int[,,]dp=new int[N,N,2]; for(int i=0; i<N; i++) { for(int j=0; j<N; j++) { for(int k=0; k<2; k++) dp[i,j,k]=0; } } // Base Case dp[0,0,0] = A[0][0]; dp[0,0,1] = A[0][0]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // Base Case if (i == 0 && j == 0) continue; // When coming from left is not // possible if (i == 0) { dp[i,j,0] = dp[i,j - 1,0] + A[i][j - 1]; dp[i,j,1] = dp[i,j - 1,1] + A[i][j - 1]; } // When coming from Left is not // possible else if (j == 0) { dp[i,j,0] = dp[i - 1,j,0] + A[i - 1][j]; dp[i,j,1] = dp[i - 1,j,1] + A[i - 1][j]; } // When Both coming from Left and Up // possible else { dp[i,j,0] = Math.Min(dp[i - 1,j,0] + A[i - 1][j], dp[i,j - 1,0] + A[i][j - 1]); dp[i,j,1] = Math.Max(dp[i - 1,j,1] + A[i - 1][j], dp[i,j - 1,1] + A[i][j - 1]); } } } // Minimum Path sum from (1, 1) to (N, N) int minimumPathSum = dp[N - 1,N - 1,0]; // Maximum Path sum from (1, 1) to (N, N) int maximumPathSum = dp[N - 1,N - 1,1]; // Number of cells in path from (1, 1) to (N, N) int numberOfCells = 2 * N - 1; // Parity of both numberOfCells from (1, 1) // to (N, N) and X should be same and X // should be in between range from // minmumPathSum and maximumPathSum if (numberOfCells % 2 == X % 2 && minimumPathSum <= X && maximumPathSum >= X) Console.Write("Yes\n"); else Console.Write("No\n"); } // Driver Code public static void Main() { // Input 1 List<List<int> > A=new List<List<int>>(); A.Add(new List<int>{ 1, 1, 1, 1 }); A.Add(new List<int>{ -1, 1, -1, -1 }); A.Add(new List<int>{ -1, 1, -1, 1 }); A.Add(new List<int>{ -1, -1, 1, 1 }); int N = A.Count; int X = 3; // Function Call isReachable(A, N, X); // Input 2 List<List<int> > A1=new List<List<int>>(); A1.Add(new List<int> { 1, 1, 1 }); A1.Add(new List<int>{ 1, 1, 1 }); A1.Add(new List<int>{ 1, 1, 1 }); int N1 = A1.Count; int X1 = 4; // Function Call isReachable(A1, N1, X1); }}// This code is contributed by poojaagarwal2. |
Javascript
// JavaScript code to implement the approach// Function to check whether paths// exists from (1, 1) to (N, N) such// that its sum is Xfunction isReachable(A, N, X) { // DP table for finding maximum and // minimum path sum let dp = new Array(N); for (let i = 0; i < N; i++) { dp[i] = new Array(N); for (let j = 0; j < N; j++) { dp[i][j] = new Array(2).fill(0); } } // Base Case dp[0][0][0] = A[0][0]; dp[0][0][1] = A[0][0]; for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { // Base Case if (i === 0 && j === 0) continue; // When coming from left is not // possible if (i === 0) { dp[i][j][0] = dp[i][j - 1][0] + A[i][j - 1]; dp[i][j][1] = dp[i][j - 1][1] + A[i][j - 1]; } // When coming from Left is not // possible else if (j === 0) { dp[i][j][0] = dp[i - 1][j][0] + A[i - 1][j]; dp[i][j][1] = dp[i - 1][j][1] + A[i - 1][j]; } // When Both coming from Left and Up // possible else { dp[i][j][0] = Math.min(dp[i - 1][j][0] + A[i - 1][j], dp[i][j - 1][0] + A[i][j - 1]); dp[i][j][1] = Math.max(dp[i - 1][j][1] + A[i - 1][j], dp[i][j - 1][1] + A[i][j - 1]); } } } // Minimum Path sum from (1, 1) to (N, N) let minimumPathSum = dp[N - 1][N - 1][0]; // Maximum Path sum from (1, 1) to (N, N) let maximumPathSum = dp[N - 1][N - 1][1]; // Number of cells in path from (1, 1) to (N, N) let numberOfCells = 2 * N - 1; // Parity of both numberOfCells from (1, 1) // to (N, N) and X should be same and X // should be in between range from // minmumPathSum and maximumPathSum if (numberOfCells % 2 === X % 2 && minimumPathSum <= X && maximumPathSum >= X) { console.log("Yes"); } else { console.log("No"); }}// Driver Codelet A = [ [ 1, 1, 1, 1 ],[ -1, 1, -1, -1 ],[ -1, 1, -1, 1 ],[ -1, -1, 1, 1 ] ]; let N = A.length; let X = 3; // Function Call isReachable(A, N, X); // Input 2 let A1= [ [ 1, 1, 1 ],[ 1, 1, 1 ],[ 1, 1, 1 ] ]; let N1 = A1.length; let X1 = 4; // Function Call isReachable(A1, N1, X1); // code by ksam24000 |
Output
Yes No
Time Complexity: O(N * N)
Auxiliary Space: O(N * N)
Related Articles :
- Introduction to Dynamic Programming – Data Structures and Algorithms Tutorials
- Introduction to Matrix or Grid – Data Structure and Algorithm Tutorials
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