Suppose A is at position (0, 0) of a 2-D grid containing ‘m’ rows and ‘n’ columns. His aim is to reach the bottom right point of this grid traveling through as minimum number of cells as possible.
Each cell of the grid contains a positive integer that defines the number of cells A can jump either in the right or the downward direction when he reaches that cell.
Find the minimum no of cells that need to be touched in order to reach bottom right corner.
Examples:
Input : 2 4 2
5 3 8
1 1 1
Output :
So following two paths exist to reach (2, 2) from (0, 0)
(0, 0) => (0, 2) => (2, 2)
(0, 0) => (2, 0) => (2, 1) => (2, 2)
Hence the output for this test case should be 3
Following is a Breadth First Search(BFS) solution of the problem:
- Think of this matrix as tree and (0, 0) as root and apply BFS using level order traversal.
- Push the coordinates and no of jumps in a queue.
- Pop the queue after every level of tree.
- Add the value at cell to the coordinates while traversing right and downward direction.
- Return no of cells touched while jumping when it reaches bottom right corner.
Approach:
- Create a function matrixJump which takes a 2D array M and two integers R1 and C1 as input.
- Create a queue q to store the number of cells and coordinates (x, y).
- Push the starting position (R1, C1) and the number of cells (1) to the queue.
- While the queue is not empty, do the following:
a. Dequeue the front element of the queue, and set x, y, and no_of_cells to the corresponding values.
b. If the current position (x, y) is the bottom right cell, return the number of cells no_of_cells.
c. Otherwise, get the value of the current position in the matrix M, and for each of the two possible moves (x+v, y) and (x, y+v), where v is the value of the current position, do the following:
i. If the move is valid (i.e., the new position is within the matrix bounds), push the new position and the updated number of cells (no_of_cells+1) to the queue. - If the bottom right cell cannot be reached, return -1.
Pseudocode:
function matrixJump(M, R1, C1): q = queue of pairs of integers and a pair of integers q.push((1, (R1, C1))) while not q.empty(): no_of_cells, (x, y) = q.front() q.pop() if x == R-1 and y == C-1: return no_of_cells v = M[x][y] if safe(x + v, y): q.push((no_of_cells+1, (x+v, y))) if safe(x, y + v): q.push((no_of_cells+1, (x, y+v))) return -1 function safe(x, y): return x < R and y < C and x >= 0 and y >= 0
Implementation:
C++
// C++ program to reach bottom right corner using// minimum jumps.#include <bits/stdc++.h>using namespace std;#define R 3#define C 3// function to check coordinates are in valid range.bool safe(int x, int y){ if (x < R && y < C && x >= 0 && y >= 0) return true; return false;}// function to return minimum no of cells to reach// bottom right cell.int matrixJump(int M[R][C], int R1, int C1){ queue<pair<int, pair<int, int> > > q; // push the no of cells and coordinates in a queue. q.push(make_pair(1, make_pair(R1, C1))); while (!q.empty()) { int x = q.front().second.first; // x coordinate int y = q.front().second.second; // y coordinate int no_of_cells = q.front().first; // no of cells q.pop(); // when it reaches bottom right return no of cells if (x == R - 1 && y == C - 1) return no_of_cells; int v = M[x][y]; if (safe(x + v, y)) q.push(make_pair(no_of_cells + 1, make_pair(x + v, y))); if (safe(x, y + v)) q.push(make_pair(no_of_cells + 1, make_pair(x, y + v))); } // when destination cannot be reached return -1;}// driver functionint main(){ int M[R][C] = { { 2, 4, 2 }, { 5, 3, 8 }, { 1, 1, 1 } }; cout << matrixJump(M, 0, 0); return 0;} |
Java
// Java program to reach bottom right corner// using minimum jumps.import java.util.*;class GFG{static int R = 3, C = 3;// function to check coordinates are in valid range.static boolean safe(int x, int y){ if (x < R && y < C && x >= 0 && y >= 0) return true; return false;}// pair classstatic class pair<T, R>{ T first; R second; pair(T t, R r) { first = t; second = r; }}// function to return minimum no of cells // to reach bottom right cell.static int matrixJump(int M[][], int R1, int C1){ Queue<pair<Integer, pair<Integer, Integer>>> q = new LinkedList<>(); // push the no of cells and coordinates in a queue. q.add(new pair(1, new pair(R1, C1))); while (q.size() > 0) { int x = q.peek().second.first; // x coordinate int y = q.peek().second.second; // y coordinate int no_of_cells = q.peek().first; // no of cells q.remove(); // when it reaches bottom right return no of cells if (x == R - 1 && y == C - 1) return no_of_cells; int v = M[x][y]; if (safe(x + v, y)) q.add(new pair(no_of_cells + 1, new pair(x + v, y))); if (safe(x, y + v)) q.add(new pair(no_of_cells + 1, new pair(x, y + v))); } // when destination cannot be reached return -1;}// Driver Codepublic static void main(String ars[]){ int M[][] = {{ 2, 4, 2 }, { 5, 3, 8 }, { 1, 1, 1 }}; System.out.print( matrixJump(M, 0, 0));}}// This code is contributed by Arnab Kundu |
Python3
# Python3 program to reach bottom # right corner using minimum jumps. R, C = 3, 3# Function to check coordinates are in valid range. def safe(x, y): if x < R and y < C and x >= 0 and y >= 0: return True return False# Function to return minimum no of # cells to reach bottom right cell. def matrixJump(M, R1, C1): q = [] # push the no of cells and coordinates in a queue. q.append((1, (R1, C1))) while len(q) != 0: x = q[0][1][0] # x coordinate y = q[0][1][1] # y coordinate no_of_cells = q[0][0] # no of cells q.pop(0) # when it reaches bottom right return no of cells if x == R - 1 and y == C - 1: return no_of_cells v = M[x][y] if safe(x + v, y): q.append((no_of_cells + 1, (x + v, y))) if safe(x, y + v): q.append((no_of_cells + 1, (x, y + v))) # when destination cannot be reached return -1# Driver code if __name__ == "__main__": M = [[2, 4, 2], [5, 3, 8], [1, 1, 1]] print(matrixJump(M, 0, 0))# This code is contributed by Rituraj Jain |
C#
// C# program to reach bottom right corner// using minimum jumps.using System;using System.Collections;using System.Collections.Generic;class GFG{ static int R = 3, C = 3;// function to check coordinates are in valid range.static Boolean safe(int x, int y){ if (x < R && y < C && x >= 0 && y >= 0) return true; return false;}// Pair classpublic class Pair<T, U> { public T first; public U second; public Pair() { } public Pair(T first, U second) { this.first = first; this.second = second; }};// function to return minimum no of cells // to reach bottom right cell.static int matrixJump(int [,]M, int R1, int C1){ Queue<Pair<int,Pair<int,int>>> q = new Queue<Pair<int,Pair<int,int>>>(); // push the no of cells and coordinates in a queue. q.Enqueue(new Pair<int, Pair<int, int>>( 1, new Pair<int, int>(R1, C1))); while (q.Count > 0) { int x = q.Peek().second.first; // x coordinate int y = q.Peek().second.second; // y coordinate int no_of_cells = q.Peek().first; // no of cells q.Dequeue(); // when it reaches bottom right return no of cells if (x == R - 1 && y == C - 1) return no_of_cells; int v = M[x, y]; if (safe(x + v, y)) q.Enqueue(new Pair<int,Pair<int,int>>(no_of_cells + 1, new Pair<int,int>(x + v, y))); if (safe(x, y + v)) q.Enqueue(new Pair<int,Pair<int,int>>(no_of_cells + 1, new Pair<int,int>(x, y + v))); } // when destination cannot be reached return -1;}// Driver Codepublic static void Main(String []ars){ int [,]M = {{ 2, 4, 2 }, { 5, 3, 8 }, { 1, 1, 1 }}; Console.Write( matrixJump(M, 0, 0));}}// This code is contributed by Arnab Kundu |
Javascript
<script>// Javascript program to reach bottom right corner// using minimum jumps.let R = 3, C = 3;// function to check coordinates are in valid range.function safe(x,y){ if (x < R && y < C && x >= 0 && y >= 0) return true; return false;}// pair classclass pair{ constructor(t,r) { this.first = t; this.second = r; }}// function to return minimum no of cells // to reach bottom right cell.function matrixJump(M,R1,C1){ let q=[]; // push the no of cells and coordinates in a queue. q.push(new pair(1, new pair(R1, C1))); while (q.length > 0) { let x = q[0].second.first; // x coordinate let y = q[0].second.second; // y coordinate let no_of_cells = q[0].first; // no of cells q.shift(); // when it reaches bottom right return no of cells if (x == R - 1 && y == C - 1) return no_of_cells; let v = M[x][y]; if (safe(x + v, y)) q.push(new pair(no_of_cells + 1, new pair(x + v, y))); if (safe(x, y + v)) q.push(new pair(no_of_cells + 1, new pair(x, y + v))); } // when destination cannot be reached return -1;}// Driver Codelet M=[[ 2, 4, 2 ],[ 5, 3, 8 ],[ 1, 1, 1 ]];document.write( matrixJump(M, 0, 0)); // This code is contributed by avanitrachhadiya2155</script> |
Output
3
Time Complexity : O(n)
Auxiliary Space : O(n)
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