Minimize cost to reach bottom-right corner of a Matrix using given operations
Last Updated :
23 Dec, 2022
Given a matrix grid[][] of size N x N, the task is to find the minimum cost required to reach the bottom right corner of the matrix from the top left corner, where the cost of moving to a new cell is [S/2] + K, where S is the score at the previous index and K is the matrix element at the current index.
Note: Here, [X] is the largest integer which is not exceeding X.
Examples:
Input: grid[][] = {{0, 3, 9, 6}, {1, 4, 4, 5}, {8, 2, 5, 4}, {1, 8, 5, 9}}
Output: 12
Explanation: One of the possible set of moves is as follows 0 -> 1 -> 4 -> 4 -> 7 -> 7 -> 12.
Input: grid[][] = {{0, 82, 2, 6, 7}, {4, 3, 1, 5, 21}, {6, 4, 20, 2, 8}, {6, 6, 64, 1, 8}, {1, 65, 1, 6, 4}}
Output: 7
Approach: Follow the steps below to solve the problem:
- Make the first element of the matrix zero.
- Traverse in the range [0, N-1]:
- Initialize a list, say moveList, and append the moves i and j.
- Initialize another list, say possibleWays, and append moveList in it.
- Initialize a list, say possibleWaysSum, initially as an empty list.
- Traverse the list possibleWays:
- Traverse the appended moveList:
- Check if the move is equal to i, then update i = i + 1.
- Otherwise, update j = j + 1.
- Initialize a variable, say tempSum. Set tempSum = tempSum + grid[i][j].
- Append tempSum in possibleWays list after the loop.
- Print the minimum cost from possibleWaysSum as the output.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
vector<vector<char> > possibleWays;
// Returns true if str[curr] does not matches with any
// of the characters after str[start]
bool shouldSwap(char str[], int start, int curr)
{
for (int i = start; i < curr; i++) {
if (str[i] == str[curr]) {
return false;
}
}
return true;
}
// Function for the swap
void swap(char str[], int i, int j)
{
char c = str[i];
str[i] = str[j];
str[j] = c;
}
// Prints all distinct permutations in str[0..n-1]
void findPermutations(char str[], int index, int n)
{
if (index >= n) {
vector<char> l;
for (int i = 0; i < n; i++)
l.push_back(str[i]);
possibleWays.push_back(l);
return;
}
for (int i = index; i < n; i++) {
// Proceed further for str[i] only if it
// doesn't match with any of the characters
// after str[index]
bool check = shouldSwap(str, index, i);
if (check) {
swap(str, index, i);
findPermutations(str, index + 1, n);
swap(str, index, i);
}
}
}
// Function to print the minimum cost
void minCost(int grid[][5], int N)
{
vector<char> moveList;
// Making top-left value 0
// implicitly to generate list of moves
grid[0][0] = 0;
vector<int> possibleWaysSum;
for (int k = 0; k < N - 1; k++) {
moveList.push_back('i');
moveList.push_back('j');
possibleWays.clear();
// Convert into set to make only unique values
int n = moveList.size();
char str[n];
for (int i = 0; i < n; i++)
str[i] = moveList[i];
// To store the unique permutation of moveList
// into the possibleWays
findPermutations(str, 0, n);
possibleWaysSum.clear();
// Traverse the list
for (vector<char> way : possibleWays) {
int i = 0, j = 0, tempSum = 0;
for (char move : way) {
if (move == 'i') {
i += 1;
}
else {
j += 1;
}
// Stores cost according to given
// conditions
tempSum = (int)(floor(tempSum / 2))
+ grid[i][j];
}
possibleWaysSum.push_back(tempSum);
}
}
// Print the minimum possible cost
int ans = possibleWaysSum[0];
for (int i = 1; i < possibleWaysSum.size(); i++)
ans = min(ans, possibleWaysSum[i]);
cout << ans;
}
// Driven Program
int main()
{
// Size of the grid
int N = 4;
// Given grid[][]
int grid[][5] = { { 0, 3, 9, 6 },
{ 1, 4, 4, 5 },
{ 8, 2, 5, 4 },
{ 1, 8, 5, 9 } };
// Function call to print the minimum
// cost to reach bottom-right corner
// from the top-left corner of the matrix
minCost(grid, N);
return 0;
}
// This code is contributed by Kingash.
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
static ArrayList<ArrayList<Character> > possibleWays;
// Returns true if str[curr] does not matches with any
// of the characters after str[start]
static boolean shouldSwap(char str[], int start,
int curr)
{
for (int i = start; i < curr; i++)
{
if (str[i] == str[curr])
{
return false;
}
}
return true;
}
// Prints all distinct permutations in str[0..n-1]
static void findPermutations(char str[], int index,
int n)
{
if (index >= n)
{
ArrayList<Character> l = new ArrayList<>();
for (char ch : str)
l.add(ch);
possibleWays.add(l);
return;
}
for (int i = index; i < n; i++) {
// Proceed further for str[i] only if it
// doesn't match with any of the characters
// after str[index]
boolean check = shouldSwap(str, index, i);
if (check) {
swap(str, index, i);
findPermutations(str, index + 1, n);
swap(str, index, i);
}
}
}
// Function for the swap
static void swap(char[] str, int i, int j)
{
char c = str[i];
str[i] = str[j];
str[j] = c;
}
// Function to print the minimum cost
static void minCost(int grid[][], int N)
{
ArrayList<Character> moveList = new ArrayList<>();
// Making top-left value 0
// implicitly to generate list of moves
grid[0][0] = 0;
ArrayList<Integer> possibleWaysSum
= new ArrayList<>();
for (int k = 0; k < N - 1; k++) {
moveList.add('i');
moveList.add('j');
possibleWays = new ArrayList<>();
// Convert into set to make only unique values
int n = moveList.size();
char str[] = new char[n];
for (int i = 0; i < n; i++)
str[i] = moveList.get(i);
// To store the unique permutation of moveList
// into the possibleWays
findPermutations(str, 0, n);
possibleWaysSum = new ArrayList<>();
// Traverse the list
for (ArrayList<Character> way : possibleWays) {
int i = 0, j = 0, tempSum = 0;
for (char move : way) {
if (move == 'i') {
i += 1;
}
else {
j += 1;
}
// Stores cost according to given
// conditions
tempSum = (int)(Math.floor(tempSum / 2))
+ grid[i][j];
}
possibleWaysSum.add(tempSum);
}
}
// Print the minimum possible cost
int ans = possibleWaysSum.get(0);
for (int i = 1; i < possibleWaysSum.size(); i++)
ans = Math.min(ans, possibleWaysSum.get(i));
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
// Size of the grid
int N = 4;
// Given grid[][]
int grid[][] = { { 0, 3, 9, 6 },
{ 1, 4, 4, 5 },
{ 8, 2, 5, 4 },
{ 1, 8, 5, 9 } };
// Function call to print the minimum
// cost to reach bottom-right corner
// from the top-left corner of the matrix
minCost(grid, N);
}
}
// This code is contributed by Kingash.
# Python3 program for the above approach
from itertools import permutations
from math import floor
# Function to print the minimum cost
def minCost(grid, N):
moveList = []
# Making top-left value 0
# implicitly to generate list of moves
grid[0][0] = 0
for i in range(N - 1):
moveList.append('i')
moveList.append('j')
# Convert into set to make only unique values
possibleWays = list(set(permutations(moveList)))
possibleWaysSum = []
# Traverse the list
for way in possibleWays:
i, j, tempSum = 0, 0, 0
for move in way:
if move == 'i':
i += 1
else:
j += 1
# Stores cost according to given conditions
tempSum = floor(tempSum/2) + grid[i][j]
possibleWaysSum.append(tempSum)
minWayIndex = possibleWaysSum.index(min(possibleWaysSum))
# Print the minimum possible cost
print(min(possibleWaysSum))
# Size of the grid
N = 4
# Given grid[][]
grid = [[0, 3, 9, 6], [1, 4, 4, 5], [8, 2, 5, 4], [1, 8, 5, 9]]
# Function call to print the minimum
# cost to reach bottom-right corner
# from the top-left corner of the matrix
minCost(grid, N)
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
static List<List<char> > possibleWays;
// Returns true if str[curr] does not matches with any
// of the chars after str[start]
static bool shouldSwap(char[] str, int start, int curr)
{
for (int i = start; i < curr; i++) {
if (str[i] == str[curr]) {
return false;
}
}
return true;
}
// Prints all distinct permutations in str[0..n-1]
static void findPermutations(char[] str, int index,
int n)
{
if (index >= n) {
List<char> l = new List<char>();
foreach(char ch in str) l.Add(ch);
possibleWays.Add(l);
return;
}
for (int i = index; i < n; i++) {
// Proceed further for str[i] only if it
// doesn't match with any of the chars
// after str[index]
bool check = shouldSwap(str, index, i);
if (check) {
swap(str, index, i);
findPermutations(str, index + 1, n);
swap(str, index, i);
}
}
}
// Function for the swap
static void swap(char[] str, int i, int j)
{
char c = str[i];
str[i] = str[j];
str[j] = c;
}
// Function to print the minimum cost
static void minCost(int[][] grid, int N)
{
List<char> moveList = new List<char>();
// Making top-left value 0
// implicitly to generate list of moves
grid[0][0] = 0;
List<int> possibleWaysSum = new List<int>();
for (int k = 0; k < N - 1; k++) {
moveList.Add('i');
moveList.Add('j');
possibleWays = new List<List<char> >();
// Convert into set to make only unique values
int n = moveList.Count;
char[] str = new char[n];
for (int i = 0; i < n; i++)
str[i] = moveList[i];
// To store the unique permutation of moveList
// into the possibleWays
findPermutations(str, 0, n);
possibleWaysSum = new List<int>();
// Traverse the list
foreach(List<char> way in possibleWays)
{
int i = 0, j = 0, tempSum = 0;
foreach(char move in way)
{
if (move == 'i') {
i += 1;
}
else {
j += 1;
}
// Stores cost according to given
// conditions
tempSum
= (int)(tempSum / 2) + grid[i][j];
}
possibleWaysSum.Add(tempSum);
}
}
// Print the minimum possible cost
int ans = possibleWaysSum[0];
for (int i = 1; i < possibleWaysSum.Count; i++)
ans = Math.Min(ans, possibleWaysSum[i]);
Console.WriteLine(ans);
}
// Driver code
public static void Main(string[] args)
{
// Size of the grid
int N = 4;
// Given grid[][]
int[][] grid
= { new[] { 0, 3, 9, 6 }, new[] { 1, 4, 4, 5 },
new[] { 8, 2, 5, 4 },
new[] { 1, 8, 5, 9 } };
// Function call to print the minimum
// cost to reach bottom-right corner
// from the top-left corner of the matrix
minCost(grid, N);
}
}
// This code is contributed by phasing17.
<script>
// Javascript program for the above approach
let possibleWays = [];
// Returns true if str[curr] does not matches with any
// of the characters after str[start]
function shouldSwap(str, start, curr) {
for (let i = start; i < curr; i++) {
if (str[i] == str[curr]) {
return false;
}
}
return true;
}
// Prints all distinct permutations in str[0..n-1]
function findPermutations(str, index, n) {
if (index >= n) {
let l = new Array();
for (let ch of str)
l.push(ch);
possibleWays.push(l);
return;
}
for (let i = index; i < n; i++) {
// Proceed further for str[i] only if it
// doesn't match with any of the characters
// after str[index]
let check = shouldSwap(str, index, i);
if (check) {
swap(str, index, i);
findPermutations(str, index + 1, n);
swap(str, index, i);
}
}
}
// Function for the swap
function swap(str, i, j) {
let c = str[i];
str[i] = str[j];
str[j] = c;
}
// Function to print the minimum cost
function minCost(grid, N) {
let moveList = new Array();
// Making top-left value 0
// implicitly to generate list of moves
grid[0][0] = 0;
let possibleWaysSum = new Array();
for (let k = 0; k < N - 1; k++) {
moveList.push('i');
moveList.push('j');
possibleWays = new Array();
// Convert into set to make only unique values
let n = moveList.length;
let str = [];
for (let i = 0; i < n; i++)
str[i] = moveList[i];
// To store the unique permutation of moveList
// into the possibleWays
findPermutations(str, 0, n);
possibleWaysSum = new Array();
// Traverse the list
for (let way of possibleWays) {
let i = 0, j = 0, tempSum = 0;
for (let move of way) {
if (move == 'i') {
i += 1;
}
else {
j += 1;
}
// Stores cost according to given
// conditions
tempSum = (Math.floor(tempSum / 2))
+ grid[i][j];
}
possibleWaysSum.push(tempSum);
}
}
// Print the minimum possible cost
let ans = possibleWaysSum[0];
for (let i = 1; i < possibleWaysSum.length; i++)
ans = Math.min(ans, possibleWaysSum[i]);
document.write(ans);
}
// Driver code
// Size of the grid
let N = 4;
// Given grid[][]
let grid = [[0, 3, 9, 6],
[1, 4, 4, 5],
[8, 2, 5, 4],
[1, 8, 5, 9]];
// Function call to print the minimum
// cost to reach bottom-right corner
// from the top-left corner of the matrix
minCost(grid, N);
// This code is contributed by Saurabh Jaiswal
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
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