You are given a 2D matrix cost[][] of dimensions m × n, where each cell represents the cost of traversing through that position. Your goal is to determine the minimum cost required to reach the bottom-right cell (m-1, n-1) starting from the top-left cell (0,0).
The total cost of a path is the sum of all cell values along the path, including both the starting and ending positions. From any cell (i, j), you can move in the following three directions:
- Right
(i, j+1) - Down
(i+1, j) - Diagonal
(i+1, j+1)
Find the minimum cost path from (0,0) to (m-1, n-1), ensuring that movement constraints are followed.
Example:
Input:
Output: 8
Explanation: The path with minimum cost is highlighted in the following figure. The path is (0, 0) –> (0, 1) –> (1, 2) –> (2, 2). The cost of the path is 8 (1 + 2 + 2 + 3).
[Naive Approach] – Using Recursion – O(3 ^ (m * n)) Time and O(1) Space
The idea is to recursively generate all possible paths from top-left cell to bottom-right cell, and find the path with minimum cost. For the recursive approach, there are three potential cases for each cell in the grid (for any cell at position (m, n)):
- The cost to reach the current cell can be calculated by moving left, i.e., from cell (m, n-1).
- The cost to reach the current cell can also be calculated by moving up, i.e., from cell (m-1, n).
- The cost can also be calculated by moving diagonally, i.e., from cell (m-1, n-1).
Mathematically, the recurrence relation will look like:
minCost(cost, m, n) = cost[m][n] + min(minCost(cost, m, n-1), minCost(cost, m-1, n), minCost(cost, m-1, n-1))
Base Cases:
- if m < 0 or n < 0 (out of bounds) minCost(cost, m, n) = INT_MAX
- minCost(cost, 0, 0) = cost[0][0] as the starting point.
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
int findMinCost(vector<vector<int>>& cost, int x, int y) {
int m = cost.size();
int n = cost[0].size();
// If indices are out of bounds, return a large value
if (x >= m || y >= m) {
return INT_MAX;
}
// Base case: bottom cell
if (x == m - 1 && y == n - 1) {
return cost[x][y];
}
// Recursively calculate minimum cost from
// all possible paths
return cost[x][y] + min({findMinCost(cost, x, y + 1),
findMinCost(cost, x + 1, y),
findMinCost(cost, x + 1, y + 1)});
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
int minCost(vector<vector<int>>& cost) {
return findMinCost(cost, 0, 0);
}
int main() {
vector<vector<int>> cost = {
{ 1, 2, 3 },
{ 4, 8, 2 },
{ 1, 5, 3 }
};
cout << minCost(cost);
return 0;
}
import java.util.*;
class GfG {
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
static int findMinCost(int[][] cost, int x, int y) {
int m = cost.length;
int n = cost[0].length;
// If indices are out of bounds, return a large value
if (x >= m || y >= n) {
return Integer.MAX_VALUE;
}
// Base case: bottom cell
if (x == m - 1 && y == n - 1) {
return cost[x][y];
}
// Recursively calculate minimum cost from
// all possible paths
return cost[x][y] + Math.min(
Math.min(findMinCost(cost, x, y + 1),
findMinCost(cost, x + 1, y)),
findMinCost(cost, x + 1, y + 1));
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
static int minCost(int[][] cost) {
return findMinCost(cost, 0, 0);
}
public static void main(String[] args) {
int[][] cost = {
{ 1, 2, 3 },
{ 4, 8, 2 },
{ 1, 5, 3 }
};
System.out.println(minCost(cost));
}
}
import sys
# Function to return the cost of the minimum cost path
# from (0,0) to (m - 1, n - 1) in a cost matrix
def findMinCost(cost, x, y):
m = len(cost)
n = len(cost[0])
# If indices are out of bounds, return a large value
if x >= m or y >= n:
return sys.maxsize
# Base case: bottom cell
if x == m - 1 and y == n - 1:
return cost[x][y]
# Recursively calculate minimum cost from
# all possible paths
return cost[x][y] + min(findMinCost(cost, x, y + 1),
findMinCost(cost, x + 1, y),
findMinCost(cost, x + 1, y + 1))
# function to find the minimum cost path
# to reach (m - 1, n - 1) from (0, 0)
def minCost(cost):
return findMinCost(cost, 0, 0)
cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
]
print(minCost(cost))
using System;
class GfG {
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
static int findMinCost(int[,] cost, int x, int y) {
int m = cost.GetLength(0);
int n = cost.GetLength(1);
// If indices are out of bounds, return a large value
if (x >= m || y >= n) {
return int.MaxValue;
}
// Base case: bottom cell
if (x == m - 1 && y == n - 1) {
return cost[x, y];
}
// Recursively calculate minimum cost from
// all possible paths
return cost[x, y] + Math.Min(
Math.Min(findMinCost(cost, x, y + 1),
findMinCost(cost, x + 1, y)),
findMinCost(cost, x + 1, y + 1));
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
static int minCost(int[,] cost) {
return findMinCost(cost, 0, 0);
}
public static void Main() {
int[,] cost = {
{ 1, 2, 3 },
{ 4, 8, 2 },
{ 1, 5, 3 }
};
Console.WriteLine(minCost(cost));
}
}
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
function findMinCost(cost, x, y) {
let m = cost.length;
let n = cost[0].length;
// If indices are out of bounds, return a large value
if (x >= m || y >= n) {
return Number.MAX_SAFE_INTEGER;
}
// Base case: bottom cell
if (x === m - 1 && y === n - 1) {
return cost[x][y];
}
// Recursively calculate minimum cost from
// all possible paths
return cost[x][y] + Math.min(
findMinCost(cost, x, y + 1),
findMinCost(cost, x + 1, y),
findMinCost(cost, x + 1, y + 1));
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
function minCost(cost) {
return findMinCost(cost, 0, 0);
}
let cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
];
console.log(minCost(cost));
[Expected Approach – 1] – Using Top-Down DP (Memoization) – O(m*n) Time and O(m*n) Space
1. Optimal Substructure:The minimum cost to reach any cell in the grid can be derived from smaller subproblems (i.e., the cost to reach neighboring cells). Specifically, the recursive relation will look like:
minCost(cost, m, n) = cost[m][n] + min(minCost(cost, m, n-1), minCost(cost, m-1, n), minCost(cost, m-1, n-1))
If the last element (i.e., cell (m – 1, n – 1)) is reached, the cost to reach it will be the value of the cell plus the minimum cost to reach one of its valid neighboring cells (left, up, or diagonal).
2. Overlapping Subproblems:
In the recursive approach, subproblems are computed multiple times. For example, when computing the minimum cost from (m, n), we may need to compute the cost for (m-1, n) multiple times. This repetition can be avoided by storing the results of subproblems in a memoization table.
memo[m][n] = cost[m][n] + min(minCost(cost, m, n-1), minCost(cost, m-1, n), minCost(cost, m-1, n-1))
Steps to implement Memoization:
- Create a 2D memo table of size (m x n) to store the computed values, initialized to -1 to indicate uncomputed subproblems.
- If we are at the last cell (m – 1, n – 1), return the value of cost[m – 1][n – 1].
- Before computing the result for any cell (m, n), check if the value at memo[m][n] is already computed. If it is, return the stored value.
- If the value is not computed, recursively calculate the minimum cost and store it in memo[m][n].
- Finally, return the value in memo[0][0] for the minimum cost to reach the target cell (m-1, n-1).
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
int findMinCost(vector<vector<int>>& cost, int x,
int y, vector<vector<int>> &memo) {
int m = cost.size();
int n = cost[0].size();
// If indices are out of bounds, return a large value
if (x >= m || y >= m) {
return INT_MAX;
}
// Base case: bottom cell
if (x == m - 1 && y == n - 1) {
return cost[x][y];
}
// Check if the result is already computed
if (memo[x][y] != -1) {
return memo[x][y];
}
// Recursively calculate minimum cost from
// all possible paths
return memo[x][y] = cost[x][y] +
min({findMinCost(cost, x, y + 1, memo),
findMinCost(cost, x + 1, y, memo),
findMinCost(cost, x + 1, y + 1, memo)});
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
int minCost(vector<vector<int>>& cost) {
int m = cost.size();
int n = cost[0].size();
// create 2d array to store the minimum cost path
vector<vector<int>> memo(m, vector<int>(n, -1));
return findMinCost(cost, 0, 0, memo);
}
int main() {
vector<vector<int>> cost = {
{ 1, 2, 3 },
{ 4, 8, 2 },
{ 1, 5, 3 }
};
cout << minCost(cost);
return 0;
}
import java.util.*;
class GfG {
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
static int findMinCost(int[][] cost, int x,
int y, int[][] memo) {
int m = cost.length;
int n = cost[0].length;
// If indices are out of bounds, return a large value
if (x >= m || y >= n) {
return Integer.MAX_VALUE;
}
// Base case: bottom cell
if (x == m - 1 && y == n - 1) {
return cost[x][y];
}
// Check if the result is already computed
if (memo[x][y] != -1) {
return memo[x][y];
}
// Recursively calculate minimum cost from
// all possible paths
return memo[x][y] = cost[x][y] +
Math.min(Math.min(findMinCost(cost, x, y + 1, memo),
findMinCost(cost, x + 1, y, memo)),
findMinCost(cost, x + 1, y + 1, memo));
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
static int minCost(int[][] cost) {
int m = cost.length;
int n = cost[0].length;
// create 2d array to store the minimum cost path
int[][] memo = new int[m][n];
for (int[] row : memo) {
Arrays.fill(row, -1);
}
return findMinCost(cost, 0, 0, memo);
}
public static void main(String[] args) {
int[][] cost = {
{ 1, 2, 3 },
{ 4, 8, 2 },
{ 1, 5, 3 }
};
System.out.println(minCost(cost));
}
}
import sys
# Function to return the cost of the minimum cost path
# from (0,0) to (m - 1, n - 1) in a cost matrix
def findMinCost(cost, x, y, memo):
m = len(cost)
n = len(cost[0])
# If indices are out of bounds, return a large value
if x >= m or y >= n:
return sys.maxsize
# Base case: bottom cell
if x == m - 1 and y == n - 1:
return cost[x][y]
# Check if the result is already computed
if memo[x][y] != -1:
return memo[x][y]
# Recursively calculate minimum cost from
# all possible paths
memo[x][y] = cost[x][y] + min(
findMinCost(cost, x, y + 1, memo),
findMinCost(cost, x + 1, y, memo),
findMinCost(cost, x + 1, y + 1, memo))
return memo[x][y]
# function to find the minimum cost path
# to reach (m - 1, n - 1) from (0, 0)
def minCost(cost):
m = len(cost)
n = len(cost[0])
# create 2d array to store the minimum cost path
memo = [[-1] * n for _ in range(m)]
return findMinCost(cost, 0, 0, memo)
cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
]
print(minCost(cost))
using System;
class GfG {
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
static int findMinCost(int[,] cost, int x,
int y, int[,] memo) {
int m = cost.GetLength(0);
int n = cost.GetLength(1);
// If indices are out of bounds, return a large value
if (x >= m || y >= n) {
return int.MaxValue;
}
// Base case: bottom cell
if (x == m - 1 && y == n - 1) {
return cost[x, y];
}
// Check if the result is already computed
if (memo[x, y] != -1) {
return memo[x, y];
}
// Recursively calculate minimum cost from
// all possible paths
return memo[x, y] = cost[x, y] +
Math.Min(Math.Min(findMinCost(cost, x, y + 1, memo),
findMinCost(cost, x + 1, y, memo)),
findMinCost(cost, x + 1, y + 1, memo));
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
static int minCost(int[,] cost) {
int m = cost.GetLength(0);
int n = cost.GetLength(1);
// create 2d array to store the minimum cost path
int[,] memo = new int[m, n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
memo[i, j] = -1;
}
}
return findMinCost(cost, 0, 0, memo);
}
public static void Main() {
int[,] cost = {
{ 1, 2, 3 },
{ 4, 8, 2 },
{ 1, 5, 3 }
};
Console.WriteLine(minCost(cost));
}
}
// Function to return the cost of the minimum cost path
// from (0,0) to (m - 1, n - 1) in a cost matrix
function findMinCost(cost, x, y, memo) {
let m = cost.length;
let n = cost[0].length;
// If indices are out of bounds, return a large value
if (x >= m || y >= n) {
return Number.MAX_SAFE_INTEGER;
}
// Base case: bottom cell
if (x === m - 1 && y === n - 1) {
return cost[x][y];
}
// Check if the result is already computed
if (memo[x][y] !== -1) {
return memo[x][y];
}
// Recursively calculate minimum cost from
// all possible paths
return memo[x][y] = cost[x][y] + Math.min(
findMinCost(cost, x, y + 1, memo),
findMinCost(cost, x + 1, y, memo),
findMinCost(cost, x + 1, y + 1, memo));
}
// function to find the minimum cost path
// to reach (m - 1, n - 1) from (0, 0)
function minCost(cost) {
let m = cost.length;
let n = cost[0].length;
// create 2d array to store the minimum cost path
let memo = Array.from({ length: m }, () => Array(n).fill(-1));
return findMinCost(cost, 0, 0, memo);
}
let cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
];
console.log(minCost(cost));
[Expected Approach – 2] – Using Bottom-Up DP (Tabulation) – O(m * n) Time and O(m * n) Space
The approach here is similar to the recursive one but instead of breaking down the problem recursively, we iteratively build up the solution by calculating it in a bottom-up manner. We create a 2D array dp of size m*n where dp[i][j] represents the minimum cost to reach cell (i, j).
Dynamic Programming Relation:
Initialize the base case: dp[0][0] = cost[0][0]
- For the first row (dp[0][j]), we can only move from the left: dp[0][j] = dp[0][j – 1] + cost[0][j] for j > 0
- For the first column (dp[i][0]), we can only move from the top: dp[i][0] = dp[i – 1][0] + cost[i][0] for i > 0
For the rest of the cells, we calculate the minimum cost by considering the minimum cost from three directions: from the left, from above, and from the diagonal:
- dp[i][j] = cost[i][j] + min(dp[i – 1][j], dp[i][j – 1], dp[i – 1][j – 1])
This formula ensures that at each step, we are choosing the minimum cost path to reach the current cell.
Below is given the implementation:
C++
// C++ implementation to find the minimum cost path
// using Tabulation (Dynamic Programming)
#include <bits/stdc++.h>
using namespace std;
int minCost(vector<vector<int>>& cost) {
int m = cost.size();
int n = cost[0].size();
vector<vector<int>> dp(m, vector<int>(n, 0));
// Initialize the base cell
dp[0][0] = cost[0][0];
// Fill the first row
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + cost[0][j];
}
// Fill the first column
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + cost[i][0];
}
// Fill the rest of the dp table
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = cost[i][j] + min({dp[i - 1][j],
dp[i][j - 1], dp[i - 1][j - 1]});
}
}
return dp[m - 1][n - 1];
}
int main() {
vector<vector<int>> cost = {
{1, 2, 3},
{4, 8, 2},
{1, 5, 3}
};
cout << minCost(cost);
return 0;
}
// Java implementation to find the minimum cost path
// using Tabulation (Dynamic Programming)
import java.util.ArrayList;
import java.util.List;
class GfG {
static int minCost(List<List<Integer>> cost) {
int m = cost.size();
int n = cost.get(0).size();
int[][] dp = new int[m][n];
// Initialize the base cell
dp[0][0] = cost.get(0).get(0);
// Fill the first row
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + cost.get(0).get(j);
}
// Fill the first column
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + cost.get(i).get(0);
}
// Fill the rest of the dp table
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = cost.get(i).get(j)
+ Math.min(dp[i - 1][j],
Math.min(dp[i][j - 1], dp[i - 1][j - 1]));
}
}
// Minimum cost to reach the bottom-right cell
return dp[m - 1][n - 1];
}
public static void main(String[] args) {
List<List<Integer>> cost = new ArrayList<>();
cost.add(List.of(1, 2, 3));
cost.add(List.of(4, 8, 2));
cost.add(List.of(1, 5, 3));
System.out.println(minCost(cost));
}
}
# Python implementation to find the minimum cost path
# using Tabulation (Dynamic Programming)
def mincost(cost):
m = len(cost)
n = len(cost[0])
dp = [[0] * n for _ in range(m)]
# Initialize the base cell
dp[0][0] = cost[0][0]
# Fill the first row
for j in range(1, n):
dp[0][j] = dp[0][j - 1] + cost[0][j]
# Fill the first column
for i in range(1, m):
dp[i][0] = dp[i - 1][0] + cost[i][0]
# Fill the rest of the dp table
for i in range(1, m):
for j in range(1, n):
dp[i][j] = cost[i][j] \
+ min(dp[i - 1][j], \
dp[i][j - 1], dp[i - 1][j - 1])
# Minimum cost to reach the
# bottom-right cell
return dp[m - 1][n - 1]
if __name__ == "__main__":
cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
]
print(mincost(cost))
// C# implementation to find the minimum cost path
// using Tabulation (Dynamic Programming)
using System;
using System.Collections.Generic;
class GfG {
static int MinCost(List<List<int>> cost) {
int m = cost.Count;
int n = cost[0].Count;
int[,] dp = new int[m, n];
// Initialize the base cell
dp[0, 0] = cost[0][0];
// Fill the first row
for (int j = 1; j < n; j++) {
dp[0, j] = dp[0, j - 1] + cost[0][j];
}
// Fill the first column
for (int i = 1; i < m; i++) {
dp[i, 0] = dp[i - 1, 0] + cost[i][0];
}
// Fill the rest of the dp table
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i, j] = cost[i][j]
+ Math.Min(dp[i - 1, j],
Math.Min(dp[i, j - 1], dp[i - 1, j - 1]));
}
}
// Minimum cost to reach the
// bottom-right cell
return dp[m - 1, n - 1];
}
static void Main() {
List<List<int>> cost = new List<List<int>> {
new List<int> { 1, 2, 3 },
new List<int> { 4, 8, 2 },
new List<int> { 1, 5, 3 }
};
Console.WriteLine(MinCost(cost));
}
}
// JavaScript implementation to find the minimum cost path
// using Tabulation (Dynamic Programming)
function minCost(cost) {
const m = cost.length;
const n = cost[0].length;
const dp = Array.from({ length: m }, () => Array(n).fill(0));
// Initialize the base cell
dp[0][0] = cost[0][0];
// Fill the first row
for (let j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + cost[0][j];
}
// Fill the first column
for (let i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + cost[i][0];
}
// Fill the rest of the dp table
for (let i = 1; i < m; i++) {
for (let j = 1; j < n; j++) {
dp[i][j] = cost[i][j]
+ Math.min(dp[i - 1][j],
dp[i][j - 1], dp[i - 1][j - 1]);
}
}
// Minimum cost to reach the bottom-right cell
return dp[m - 1][n - 1];
}
const cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
];
console.log(minCost(cost));
[Optimal Approach] – Using Space Optimized DP – O(m * n) Time and O(n) Space
In the previous approach, we used a 2D dp table to store the minimum cost at each cell. However, we can optimize the space complexity by observing that for calculating the current state, we only need the values from the previous row. Therefore, there is no need to store the entire dp table, and we can optimize the space to O(n) by only keeping track of the current and previous rows.
Below is given the implementation:
C++
// C++ implementation to find the minimum cost path
// using Space Optimization
#include <bits/stdc++.h>
using namespace std;
int minCost(vector<vector<int>> &cost) {
int m = cost.size();
int n = cost[0].size();
// 1D dp array to store the minimum cost
// of the current row
vector<int> dp(n, 0);
// Initialize the base cell
dp[0] = cost[0][0];
// Fill the first row
for (int j = 1; j < n; j++) {
dp[j] = dp[j - 1] + cost[0][j];
}
// Fill the rest of the rows
for (int i = 1; i < m; i++) {
// Store the previous value of dp[j-1]
// (for diagonal handling)
int prev = dp[0];
// Update the first column (only depends on
// the previous row)
dp[0] = dp[0] + cost[i][0];
for (int j = 1; j < n; j++) {
// Store the current dp[j] before updating it
int temp = dp[j];
// Update dp[j] using the minimum of the
// top, left, and diagonal cells
dp[j] = cost[i][j] + min({dp[j], dp[j - 1], prev});
// Update prev to be the old dp[j] for the
// diagonal calculation in the next iteration
prev = temp;
}
}
// The last cell contains the
// minimum cost path
return dp[n - 1];
}
int main() {
vector<vector<int>> cost = {{1, 2, 3}, {4, 8, 2}, {1, 5, 3}};
cout << minCost(cost) << endl;
return 0;
}
// Java implementation to find the minimum cost path
// using Space Optimization
import java.util.*;
class GfG {
static int minCost(int[][] cost) {
int m = cost.length;
int n = cost[0].length;
// 1D dp array to store the minimum cost of the
// current row
int[] dp = new int[n];
// Initialize the base cell
dp[0] = cost[0][0];
// Fill the first row
for (int j = 1; j < n; j++) {
dp[j] = dp[j - 1] + cost[0][j];
}
// Fill the rest of the rows
for (int i = 1; i < m; i++) {
// Store the previous value of dp[j-1] (for
// diagonal handling)
int prev = dp[0];
// Update the first column (only depends on the
// previous row)
dp[0] = dp[0] + cost[i][0];
for (int j = 1; j < n; j++) {
// Store the current dp[j] before updating
// it
int temp = dp[j];
// Update dp[j] using the minimum of the
// top, left, and diagonal cells
dp[j]
= cost[i][j]
+ Math.min(dp[j],
Math.min(dp[j - 1], prev));
// Update prev to be the old dp[j] for the
// diagonal calculation in the next
// iteration
prev = temp;
}
}
// The last cell contains the minimum cost path
return dp[n - 1];
}
public static void main(String[] args) {
int[][] cost
= { { 1, 2, 3 }, { 4, 8, 2 }, { 1, 5, 3 } };
System.out.println(minCost(cost));
}
}
# Python implementation to find the minimum cost path
# using Space Optimization
def minCost(cost):
m = len(cost)
n = len(cost[0])
# 1D dp array to store the minimum
# cost of the current row
dp = [0] * n
# Initialize the base cell
dp[0] = cost[0][0]
# Fill the first row
for j in range(1, n):
dp[j] = dp[j - 1] + cost[0][j]
# Fill the rest of the rows
for i in range(1, m):
# Store the previous value of dp[j-1]
# (for diagonal handling)
prev = dp[0]
# Update the first column (only depends
# on the previous row)
dp[0] = dp[0] + cost[i][0]
for j in range(1, n):
# Store the current dp[j] before
# updating it
temp = dp[j]
# Update dp[j] using the minimum of the top,
# left, and diagonal cells
dp[j] = cost[i][j] + min(dp[j], dp[j - 1], prev)
# Update prev to be the old dp[j] for the
# diagonal calculation in the next iteration
prev = temp
# The last cell contains the minimum
# cost path
return dp[n - 1]
cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
]
print(minCost(cost))
// C# implementation to find the minimum cost path
// using Space Optimization
using System;
class GfG {
static int MinCost(int[, ] cost) {
int m = cost.GetLength(0);
int n = cost.GetLength(1);
// 1D dp array to store the minimum cost of the
// current row
int[] dp = new int[n];
// Initialize the base cell
dp[0] = cost[0, 0];
// Fill the first row
for (int j = 1; j < n; j++) {
dp[j] = dp[j - 1] + cost[0, j];
}
// Fill the rest of the rows
for (int i = 1; i < m; i++) {
// Store the previous value of dp[j-1] (for
// diagonal handling)
int prev = dp[0];
// Update the first column (only depends on the
// previous row)
dp[0] = dp[0] + cost[i, 0];
for (int j = 1; j < n; j++) {
// Store the current dp[j] before updating
// it
int temp = dp[j];
// Update dp[j] using the minimum of the
// top, left, and diagonal cells
dp[j]
= cost[i, j]
+ Math.Min(dp[j],
Math.Min(dp[j - 1], prev));
// Update prev to be the old dp[j] for the
// diagonal calculation in the next
// iteration
prev = temp;
}
}
// The last cell contains the minimum cost path
return dp[n - 1];
}
static void Main(string[] args) {
int[, ] cost
= { { 1, 2, 3 }, { 4, 8, 2 }, { 1, 5, 3 } };
Console.WriteLine(MinCost(cost));
}
}
// JavaScript implementation to find the minimum cost path
// using Space Optimization
function minCost(cost) {
const m = cost.length;
const n = cost[0].length;
// 1D dp array to store the minimum cost of the current
// row
let dp = new Array(n);
// Initialize the base cell
dp[0] = cost[0][0];
// Fill the first row
for (let j = 1; j < n; j++) {
dp[j] = dp[j - 1] + cost[0][j];
}
// Fill the rest of the rows
for (let i = 1; i < m; i++) {
// Store the previous value of dp[j-1] (for diagonal
// handling)
let prev = dp[0];
// Update the first column (only depends on the
// previous row)
dp[0] = dp[0] + cost[i][0];
for (let j = 1; j < n; j++) {
// Store the current dp[j] before updating it
let temp = dp[j];
// Update dp[j] using the minimum of the top,
// left, and diagonal cells
dp[j] = cost[i][j]
+ Math.min(dp[j],
Math.min(dp[j - 1], prev));
// Update prev to be the old dp[j] for the
// diagonal calculation in the next iteration
prev = temp;
}
}
// The last cell contains the minimum cost path
return dp[n - 1];
}
const cost = [ [ 1, 2, 3 ], [ 4, 8, 2 ], [ 1, 5, 3 ] ];
console.log(minCost(cost));
[Alternate Less Efficient Approach] – Using Dijkstra’s Algorithm – O((m * n) * log(m * n)) Time and O(m * n) Space
The idea is to apply Dijskra’s Algorithm to find the minimum cost path from the top-left to the bottom-right corner of the grid. Each cell is treated as a node and each move between adjacent cells has a cost. We use a min-heap to always expand the least costly path first.
C++
// C++ implementation to find minimum cost path
// using Dijstra's Algoritms
#include <bits/stdc++.h>
using namespace std;
int minCost(vector<vector<int>> &cost) {
int m = cost.size();
int n = cost[0].size();
// Directions for moving down, right, and diagonal
vector<pair<int, int>> directions =
{{1, 0}, {0, 1}, {1, 1}};
// Min-heap (priority queue) for Dijkstra's algorithm
priority_queue<vector<int>,
vector<vector<int>>, greater<vector<int>>> pq;
// Distance matrix to store the minimum
// cost to reach each cell
vector<vector<int>> dist(m, vector<int>(n, INT_MAX));
dist[0][0] = cost[0][0];
pq.push({cost[0][0], 0, 0});
// Prim's algorithm
while (!pq.empty()) {
vector<int> curr = pq.top();
pq.pop();
int x = curr[1];
int y = curr[2];
// If we reached the bottom-right
// corner, return the cost
if (x == m - 1 && y == n - 1) {
return dist[x][y];
}
// Explore the neighbors
for (auto &dir : directions) {
int newX = x + dir.first;
int newY = y + dir.second;
// Ensure the new cell is within bounds
if (newX < m && newY < n) {
// Relaxation step
if (dist[newX][newY] > dist[x][y] + cost[newX][newY]) {
dist[newX][newY] = dist[x][y] + cost[newX][newY];
pq.push({dist[newX][newY], newX, newY});
}
}
}
}
return dist[m - 1][n - 1];
}
int main() {
vector<vector<int>> cost = {
{1, 2, 3},
{4, 8, 2},
{1, 5, 3}
};
cout << minCost(cost);
return 0;
}
import java.util.*;
class GfG {
int minCost(int[][] cost) {
int m = cost.length;
int n = cost[0].length;
// Directions for moving down, right, and diagonal
int[][] directions = {{1, 0}, {0, 1}, {1, 1}};
// Min-heap (priority queue) for Dijkstra's algorithm
PriorityQueue<int[]> pq =
new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
// Distance matrix to store the minimum
// cost to reach each cell
int[][] dist = new int[m][n];
for (int[] row : dist) {
Arrays.fill(row, Integer.MAX_VALUE);
}
dist[0][0] = cost[0][0];
pq.add(new int[]{cost[0][0], 0, 0});
// Prim's algorithm
while (!pq.isEmpty()) {
int[] curr = pq.poll();
int x = curr[1];
int y = curr[2];
// If we reached the bottom-right
// corner, return the cost
if (x == m - 1 && y == n - 1) {
return dist[x][y];
}
// Explore the neighbors
for (int[] dir : directions) {
int newX = x + dir[0];
int newY = y + dir[1];
// Ensure the new cell is within bounds
if (newX < m && newY < n) {
// Relaxation step
if (dist[newX][newY] > dist[x][y] + cost[newX][newY]) {
dist[newX][newY] = dist[x][y] + cost[newX][newY];
pq.add(new int[]{dist[newX][newY], newX, newY});
}
}
}
}
return dist[m - 1][n - 1];
}
public static void main(String[] args) {
int[][] cost = {
{1, 2, 3},
{4, 8, 2},
{1, 5, 3}
};
GfG obj = new GfG();
System.out.println(obj.minCost(cost));
}
}
import heapq
def minCost(cost):
m = len(cost)
n = len(cost[0])
# Directions for moving down, right, and diagonal
directions = [(1, 0), (0, 1), (1, 1)]
# Min-heap (priority queue) for Dijkstra's algorithm
pq = []
# Distance matrix to store the minimum
# cost to reach each cell
dist = [[float('inf')] * n for _ in range(m)]
dist[0][0] = cost[0][0]
heapq.heappush(pq, (cost[0][0], 0, 0))
# Prim's algorithm
while pq:
curr = heapq.heappop(pq)
x = curr[1]
y = curr[2]
# If we reached the bottom-right
# corner, return the cost
if x == m - 1 and y == n - 1:
return dist[x][y]
# Explore the neighbors
for dx, dy in directions:
newX = x + dx
newY = y + dy
# Ensure the new cell is within bounds
if newX < m and newY < n:
# Relaxation step
if dist[newX][newY] > dist[x][y] + cost[newX][newY]:
dist[newX][newY] = dist[x][y] + cost[newX][newY]
heapq.heappush(pq, (dist[newX][newY], newX, newY))
return dist[m - 1][n - 1]
cost = [
[1, 2, 3],
[4, 8, 2],
[1, 5, 3]
]
print(minCost(cost))
using System;
using System.Collections.Generic;
class GfG {
int minCost(int[,] cost) {
int m = cost.GetLength(0);
int n = cost.GetLength(1);
// Directions for moving down, right, and diagonal
int[,] directions = { {1, 0}, {0, 1}, {1, 1} };
// Min-heap (priority queue) for Dijkstra's algorithm
SortedSet<(int, int, int)> pq = new SortedSet<(int, int, int)>
(Comparer<(int, int, int)>.Create((a, b) =>
a.Item1 == b.Item1 ? (a.Item2 == b.Item2 ?
a.Item3 - b.Item3 : a.Item2 - b.Item2) : a.Item1 - b.Item1));
// Distance matrix to store the minimum
// cost to reach each cell
int[,] dist = new int[m, n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
dist[i, j] = int.MaxValue;
}
}
dist[0, 0] = cost[0, 0];
pq.Add((cost[0, 0], 0, 0));
// Prim's algorithm
while (pq.Count > 0) {
var curr = pq.Min;
pq.Remove(curr);
int x = curr.Item2;
int y = curr.Item3;
// If we reached the bottom-right
// corner, return the cost
if (x == m - 1 && y == n - 1) {
return dist[x, y];
}
// Explore the neighbors
for (int i = 0; i < 3; i++) {
int newX = x + directions[i, 0];
int newY = y + directions[i, 1];
// Ensure the new cell is within bounds
if (newX < m && newY < n) {
// Relaxation step
if (dist[newX, newY] > dist[x, y] + cost[newX, newY]) {
dist[newX, newY] = dist[x, y] + cost[newX, newY];
pq.Add((dist[newX, newY], newX, newY));
}
}
}
}
return dist[m - 1, n - 1];
}
public static void Main() {
int[,] cost = {
{1, 2, 3},
{4, 8, 2},
{1, 5, 3}
};
GfG obj = new GfG();
Console.WriteLine(obj.minCost(cost));
}
}
function minCost(cost) {
let m = cost.length;
let n = cost[0].length;
// Directions for moving down, right, and diagonal
let directions = [[1, 0], [0, 1], [1, 1]];
// Min-heap (priority queue) for Dijkstra's algorithm
let pq = [];
// Distance matrix to store the minimum
// cost to reach each cell
let dist = Array.from({ length: m },
() => Array(n).fill(Infinity));
dist[0][0] = cost[0][0];
pq.push([cost[0][0], 0, 0]);
// Prim's algorithm
while (pq.length > 0) {
pq.sort((a, b) => a[0] - b[0]);
let [currCost, x, y] = pq.shift();
// If we reached the bottom-right
// corner, return the cost
if (x === m - 1 && y === n - 1) {
return dist[x][y];
}
// Explore the neighbors
for (let [dx, dy] of directions) {
let newX = x + dx;
let newY = y + dy;
// Ensure the new cell is within bounds
if (newX < m && newY < n) {
// Relaxation step
if (dist[newX][newY] > dist[x][y] + cost[newX][newY]) {
dist[newX][newY] = dist[x][y] + cost[newX][newY];
pq.push([dist[newX][newY], newX, newY]);
}
}
}
}
return dist[m - 1][n - 1];
}
console.log(minCost([[1, 2, 3], [4, 8, 2], [1, 5, 3]]));
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Weighted Job Scheduling
Given a 2D array jobs[][] of order n*3, where each element jobs[i] defines start time, end time, and the profit associated with the job. The task is to find the maximum profit you can take such that there are no two jobs with overlapping time ranges. Note: If the job ends at time X, it is allowed to
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Count Derangements (Permutation such that no element appears in its original position)
A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of [0, 1, 2, 3] is [2, 3, 1, 0].Given a number n, find the total number of Derangements of a set of n elements. Examples : Input: n = 2Output: 1Explanation: For two balls [
12 min read
Minimum insertions to form a palindrome
Given string str, the task is to find the minimum number of characters to be inserted to convert it to a palindrome. Examples: Input: str = abcdOutput: 3Explanation: Here we can append 3 characters in the beginning, and the resultant string will be a palindrome ("dcbabcd"). Input: str = abaOutput: 0
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Ways to arrange Balls such that adjacent balls are of different types
There are 'p' balls of type P, 'q' balls of type Q and 'r' balls of type R. Using the balls we want to create a straight line such that no two balls of the same type are adjacent.Examples : Input: p = 1, q = 1, r = 0Output: 2Explanation: There are only two arrangements PQ and QP Input: p = 1, q = 1,
15+ min read