Given a binary matrix of size NxN where 1 denotes that the number i can be converted to j, and 0 denotes it cannot be converted to. Also given are two numbers X(<N)and Y(<N), the task is to find the minimum number of steps required to convert the number X to Y. If there is no such way possible, print -1.
Examples:
Input:
{{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 1, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}
{ 0, 0, 0, 1, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}}
X = 2, Y = 3
Output: 2
Convert 2 -> 4 -> 3, which is the minimum way possible.
Input:
{{ 0, 0, 0, 0}
{ 0, 0, 0, 1}
{ 0, 0, 0, 0}
{ 0, 1, 0, 0}}
X = 1, Y = 2
Output: -1
Approach: This problem is a variant of the Floyd-warshall algorithm where there is an edge of weight 1 between i and j i.e. mat[i][j]==1, else they don’t have an edge and we can assign edges as infinity as we do in Floyd-Warshall. Find the solution matrix and return dp[i][j] if it is not infinite. Return -1 if it is infinite which means there is no path possible between them.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;#define INF 99999#define size 10int findMinimumSteps(int mat[size][size], int x, int y, int n){ // dist[][] will be the output matrix that // will finally have the shortest // distances between every pair of numbers int dist[n][n], i, j, k; // Initially same as mat for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { if (mat[i][j] == 0) dist[i][j] = INF; else dist[i][j] = 1; if (i == j) dist[i][j] = 1; } } // Add all numbers one by one to the set // of intermediate numbers. Before start of // an iteration, we have shortest distances // between all pairs of numbers such that the // shortest distances consider only the numbers // in set {0, 1, 2, .. k-1} as intermediate numbers. // After the end of an iteration, vertex no. k is // added to the set of intermediate numbers and // the set becomes {0, 1, 2, .. k} for (k = 0; k < n; k++) { // Pick all numbers as source one by one for (i = 0; i < n; i++) { // Pick all numbers as destination for the // above picked source for (j = 0; j < n; j++) { // If number k is on the shortest path from // i to j, then update the value of dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]) dist[i][j] = dist[i][k] + dist[k][j]; } } } // If no path if (dist[x][y] < INF) return dist[x][y]; else return -1;}// Driver Codeint main(){ int mat[size][size] = { { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 }, { 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } }; int x = 2, y = 3; cout << findMinimumSteps(mat, x, y, size);} |
Java
// Java implementation of the above approachclass GFG{ static int INF=99999; static int findMinimumSteps(int mat[][], int x, int y, int n) { // dist[][] will be the output matrix that // will finally have the shortest // distances between every pair of numbers int i, j, k; int [][] dist= new int[n][n]; // Initially same as mat for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { if (mat[i][j] == 0) dist[i][j] = INF; else dist[i][j] = 1; if (i == j) dist[i][j] = 1; } } // Add all numbers one by one to the set // of intermediate numbers. Before start of // an iteration, we have shortest distances // between all pairs of numbers such that the // shortest distances consider only the numbers // in set {0, 1, 2, .. k-1} as intermediate numbers. // After the end of an iteration, vertex no. k is // added to the set of intermediate numbers and // the set becomes {0, 1, 2, .. k} for (k = 0; k < n; k++) { // Pick all numbers as source one by one for (i = 0; i < n; i++) { // Pick all numbers as destination for the // above picked source for (j = 0; j < n; j++) { // If number k is on the shortest path from // i to j, then update the value of dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]) dist[i][j] = dist[i][k] + dist[k][j]; } } } // If no path if (dist[x][y] < INF) return dist[x][y]; else return -1; } // Driver Code public static void main(String []args) { int [][] mat = { { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 }, { 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } }; int x = 2, y = 3; int size=mat.length; System.out.println( findMinimumSteps(mat, x, y, size)); }}// This code is contributed by ihritik |
Python3
# Python3 implementation of the above approachINF = 99999size = 10def findMinimumSteps(mat, x, y, n): # dist[][] will be the output matrix # that will finally have the shortest # distances between every pair of numbers dist = [[0 for i in range(n)] for i in range(n)] i, j, k = 0, 0, 0 # Initially same as mat for i in range(n): for j in range(n): if (mat[i][j] == 0): dist[i][j] = INF else: dist[i][j] = 1 if (i == j): dist[i][j] = 1 # Add all numbers one by one to the set # of intermediate numbers. Before start # of an iteration, we have shortest distances # between all pairs of numbers such that the # shortest distances consider only the numbers # in set {0, 1, 2, .. k-1} as intermediate # numbers. After the end of an iteration, vertex # no. k is added to the set of intermediate # numbers and the set becomes {0, 1, 2, .. k} for k in range(n): # Pick all numbers as source one by one for i in range(n): # Pick all numbers as destination # for the above picked source for j in range(n): # If number k is on the shortest path from # i to j, then update the value of dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]): dist[i][j] = dist[i][k] + dist[k][j] # If no path if (dist[x][y] < INF): return dist[x][y] else: return -1# Driver Codemat = [[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ], [ 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ]]x, y = 2, 3print(findMinimumSteps(mat, x, y, size))# This code is contributed by Mohit kumar 29 |
C#
// C# implementation of the above approach using System;class GFG { static int INF=99999; static int findMinimumSteps(int [,]mat, int x, int y, int n) { // dist[][] will be the output matrix that // will finally have the shortest // distances between every pair of numbers int i, j, k; int [,] dist= new int[n,n]; // Initially same as mat for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { if (mat[i,j] == 0) dist[i,j] = INF; else dist[i,j] = 1; if (i == j) dist[i,j] = 1; } } // Add all numbers one by one to the set // of intermediate numbers. Before start of // an iteration, we have shortest distances // between all pairs of numbers such that the // shortest distances consider only the numbers // in set {0, 1, 2, .. k-1} as intermediate numbers. // After the end of an iteration, vertex no. k is // added to the set of intermediate numbers and // the set becomes {0, 1, 2, .. k} for (k = 0; k < n; k++) { // Pick all numbers as source one by one for (i = 0; i < n; i++) { // Pick all numbers as destination for the // above picked source for (j = 0; j < n; j++) { // If number k is on the shortest path from // i to j, then update the value of dist[i][j] if (dist[i,k] + dist[k,j] < dist[i,j]) dist[i,j] = dist[i,k] + dist[k,j]; } } } // If no path if (dist[x,y] < INF) return dist[x,y]; else return -1; } // Driver Code public static void Main() { int [,] mat = { { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 }, { 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } }; int x = 2, y = 3; int size = mat.GetLength(0) ; Console.WriteLine( findMinimumSteps(mat, x, y, size)); } // This code is contributed by Ryuga} |
Javascript
<script>// JavaScript implementation of the above approach var INF=99999; function findMinimumSteps(mat , x , y , n) { // dist will be the output matrix that // will finally have the shortest // distances between every pair of numbers var i, j, k; var dist= Array(n).fill().map(()=>Array(n).fill(0)); // Initially same as mat for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { if (mat[i][j] == 0) dist[i][j] = INF; else dist[i][j] = 1; if (i == j) dist[i][j] = 1; } } // Add all numbers one by one to the set // of intermediate numbers. Before start of // an iteration, we have shortest distances // between all pairs of numbers such that the // shortest distances consider only the numbers // in set {0, 1, 2, .. k-1} as intermediate numbers. // After the end of an iteration, vertex no. k is // added to the set of intermediate numbers and // the set becomes {0, 1, 2, .. k} for (k = 0; k < n; k++) { // Pick all numbers as source one by one for (i = 0; i < n; i++) { // Pick all numbers as destination for the // above picked source for (j = 0; j < n; j++) { // If number k is on the // shortest path from // i to j, then update the // value of dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]) dist[i][j] = dist[i][k] + dist[k][j]; } } } // If no path if (dist[x][y] < INF) return dist[x][y]; else return -1; } // Driver Code var mat = [[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ], [ 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ] ]; var x = 2, y = 3; var size=mat.length; document.write( findMinimumSteps(mat, x, y, size));// This code contributed by Rajput-Ji</script> |
Output
2
Complexity Analysis:
- Time Complexity: O(N3), as we are using nested loops to traverse N3 times.
- Auxiliary Space: O(N2), as we are using extra space for the dist matrix.
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