Maximum path sum in a triangle.
Last Updated :
12 Sep, 2023
We have given numbers in form of a triangle, by starting at the top of the triangle and moving to adjacent numbers on the row below, find the maximum total from top to bottom.
Examples :
Input :
3
7 4
2 4 6
8 5 9 3
Output : 23
Explanation : 3 + 7 + 4 + 9 = 23
Input :
8
-4 4
2 2 6
1 1 1 1
Output : 19
Explanation : 8 + 4 + 6 + 1 = 19
Method 1: We can go through the brute force by checking every possible path but that is much time taking so we should try to solve this problem with the help of dynamic programming which reduces the time complexity.
Implementation of Recursive Approach:
C++
#include<bits/stdc++.h>
using namespace std;
#define N 3
int maxPathSum(int tri[][N], int i, int j, int row, int col){
if(j == col ){
return 0;
}
if(i == row-1 ){
return tri[i][j] ;
}
return tri[i][j] + max(maxPathSum(tri, i+1, j, row, col),
maxPathSum(tri, i+1, j+1, row, col)) ;
}
int main()
{
int tri[N][N] = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
cout << maxPathSum(tri, 0, 0, 3, 3);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int N = 3;
public static int maxPathSum(int tri[][], int i, int j,
int row, int col)
{
if (j == col) {
return 0;
}
if (i == row - 1) {
return tri[i][j];
}
return tri[i][j]
+ Math.max(
maxPathSum(tri, i + 1, j, row, col),
maxPathSum(tri, i + 1, j + 1, row, col));
}
public static void main(String[] args)
{
int tri[][]
= { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
System.out.print(maxPathSum(tri, 0, 0, 3, 3));
}
}
|
Python3
N = 3
def maxPathSum(tri, i, j, row, col):
if(j == col ):
return 0
if(i == row-1 ):
return tri[i][j]
return tri[i][j] + max(maxPathSum(tri, i+1, j, row, col),
maxPathSum(tri, i+1, j+1, row, col))
tri = [ [1, 0, 0],[4, 8, 0],[1, 5, 3] ]
print(maxPathSum(tri, 0, 0, 3, 3))
|
C#
using System;
public class GFG
{
static int N = 3;
public static int max(int a, int b){
if(a > b) return a;
return b;
}
public static int maxPathSum(int[,] tri, int i, int j, int row, int col)
{
if (j == col) {
return 0;
}
if (i == row - 1) {
return tri[i,j];
}
return tri[i,j]
+ max(
maxPathSum(tri, i + 1, j, row, col),
maxPathSum(tri, i + 1, j + 1, row, col)
);
}
public static void Main(string[] args)
{
int[,] tri = {{ 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
Console.WriteLine(maxPathSum(tri, 0, 0, 3, 3));
}
}
|
Javascript
<script>
const N = 3
function maxPathSum(tri, i, j, row, col){
if(j == col ){
return 0;
}
if(i == row-1 ){
return tri[i][j] ;
}
return tri[i][j] + Math.max(maxPathSum(tri, i+1, j, row, col),
maxPathSum(tri, i+1, j+1, row, col)) ;
}
let tri = [ [1, 0, 0],[4, 8, 0],[1, 5, 3] ];
document.write(maxPathSum(tri, 0, 0, 3, 3));
</script>
|
Complexity Analysis:
- Time Complexity: O(2N*N)
- Auxiliary Space: O(N)
If we should left shift every element and put 0 at each empty position to make it a regular matrix, then our problem looks like minimum cost path.
So, after converting our input triangle elements into a regular matrix we should apply the dynamic programming concept to find the maximum path sum.
Method 2: DP Top-Down
Since there are overlapping subproblems, we can avoid the repeated work done in method 1 by storing the min-cost path calculated so far using top-down approach
C++
#include<bits/stdc++.h>
using namespace std;
#define N 3
int maxPathSum(int tri[][N], int i, int j, int row, int col, vector<vector<int>> &dp){
if(j == col ){
return 0;
}
if(i == row-1 ){
return tri[i][j] ;
}
if(dp[i][j] != -1){
return dp[i][j] ;
}
return dp[i][j] = tri[i][j] + max(maxPathSum(tri, i+1, j, row, col, dp),
maxPathSum(tri, i+1, j+1, row, col, dp)) ;
}
int main()
{
int tri[N][N] = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
vector<vector<int>> dp(N, vector<int>(N, -1) ) ;
cout << maxPathSum(tri, 0, 0, N, N, dp);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int maxPathSum(int[][] tri, int i, int j,
int row, int col, int[][] dp)
{
if (j == col) {
return 0;
}
if (i == row - 1) {
return tri[i][j];
}
if (dp[i][j] != -1) {
return dp[i][j];
}
return dp[i][j]
= tri[i][j]
+ Math.max(
maxPathSum(tri, i + 1, j, row, col, dp),
maxPathSum(tri, i + 1, j + 1, row, col,
dp));
}
public static void main(String[] args)
{
int n = 3;
int[][] tri
= { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
int[][] dp = new int[n][n];
for (int[] row : dp) {
Arrays.fill(row, -1);
}
System.out.print(maxPathSum(tri, 0, 0, n, n, dp));
}
}
|
Python3
N = 3
def maxPathSum(tri, i, j, row, col, dp):
if(j == col):
return 0
if(i == row-1):
return tri[i][j]
if(dp[i][j] != -1):
return -1
dp[i][j] = tri[i][j] + max(maxPathSum(tri, i+1, j, row, col, dp),
maxPathSum(tri, i+1, j+1, row, col, dp))
return dp[i][j]
tri = [ [1, 0, 0],
[4, 8, 0],
[1, 5, 3] ]
dp = [[-1 for i in range(N)]for j in range(N)]
print(maxPathSum(tri, 0, 0, N, N, dp))
|
C#
using System;
class GFG
{
static int maxPathSum(int[, ] tri, int i, int j,
int row, int col, int[, ] dp)
{
if (j == col) {
return 0;
}
if (i == row - 1) {
return tri[i, j];
}
if (dp[i, j] != -1) {
return dp[i, j];
}
return dp[i, j]
= tri[i, j]
+ Math.Max(
maxPathSum(tri, i + 1, j, row, col, dp),
maxPathSum(tri, i + 1, j + 1, row, col,
dp));
}
static void Main()
{
int[, ] tri
= { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
int N = 3;
int[, ] dp = new int[N, N];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
dp[i, j] = -1;
Console.Write(maxPathSum(tri, 0, 0, N, N, dp));
}
}
|
Javascript
<script>
b
const N = 3
function maxPathSum(tri, i, j, row, col, dp){
if(j == col)
return 0
if(i == row-1)
return tri[i][j]
if(dp[i][j] != -1)
return -1
dp[i][j] = tri[i][j] + Math.max(maxPathSum(tri, i+1, j, row, col, dp),
maxPathSum(tri, i+1, j+1, row, col, dp))
return dp[i][j]
}
let tri = [ [1, 0, 0],
[4, 8, 0],
[1, 5, 3] ]
let dp = new Array(N).fill(-1).map(()=>new Array(N).fill(-1))
document.write(maxPathSum(tri, 0, 0, N, N, dp),"</br>")
</script>
|
Complexity Analysis:
- Time Complexity: O(m*n) where m = no of rows and n = no of columns
- Auxiliary Space: O(n2)
Method 3: DP(Bottom – UP)
Since there are overlapping subproblems, we can avoid the repeated work done in method 1 by storing the min-cost path calculated so far using the bottom-up approach thus reducing stack space
C++
#include<bits/stdc++.h>
using namespace std;
#define N 3
int maxPathSum(int tri[][N], int n, vector<vector<int>> &dp)
{
for(int j = 0; j < n; j++ ){
dp[n-1][j] = tri[n-1][j] ;
}
for(int i = n-2; i >= 0; i--){
for(int j = i; j >= 0; j-- ){
dp[i][j] = tri[i][j] + max(dp[i+1][j] , dp[i+1][j+1]) ;
}
}
return dp[0][0] ;
}
int main()
{
int tri[N][N] = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
vector<vector<int>> dp(N, vector<int>(N, -1) ) ;
cout << maxPathSum(tri, N, dp);
return 0;
}
|
Java
public class Main
{
static int maxPathSum(int[][] tri, int n, int[][] dp)
{
for (int j = 0; j < n; j++) {
dp[n - 1][j] = tri[n - 1][j];
}
for (int i = n - 2; i >= 0; i--) {
for (int j = i; j >= 0; j--) {
dp[i][j] = tri[i][j]
+ Math.max(dp[i + 1][j],
dp[i + 1][j + 1]);
}
}
return dp[0][0];
}
public static void main(String[] args)
{
int N = 3;
int[][] tri
= { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
int[][] dp = new int[N][N];
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
dp[i][j] = -1;
}
}
System.out.println(maxPathSum(tri, N, dp));
}
}
|
Python3
N = 3
def maxPathSum(tri, n, dp):
for j in range(n):
dp[n-1][j] = tri[n-1][j]
for i in range(n-2, -1, -1):
for j in range(i, -1, -1):
dp[i][j] = tri[i][j] + max(dp[i+1][j], dp[i+1][j+1])
return dp[0][0]
if __name__ == '__main__':
tri = [[1, 0, 0],
[4, 8, 0],
[1, 5, 3]]
dp = [[-1 for j in range(N)] for i in range(N)]
print(maxPathSum(tri, N, dp))
|
C#
using System;
public class GFG {
static int maxPathSum(int[, ] tri, int n, int[, ] dp)
{
for (int j = 0; j < n; j++) {
dp[n - 1, j] = tri[n - 1, j];
}
for (int i = n - 2; i >= 0; i--) {
for (int j = i; j >= 0; j--) {
dp[i, j] = tri[i, j]
+ Math.Max(dp[i + 1, j],
dp[i + 1, j + 1]);
}
}
return dp[0, 0];
}
public static void Main(string[] args)
{
int N = 3;
int[, ] tri = new int[, ] { { 1, 0, 0 },
{ 4, 8, 0 },
{ 1, 5, 3 } };
int[, ] dp = new int[N, N];
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
dp[i, j] = -1;
}
}
Console.WriteLine(maxPathSum(tri, N, dp));
}
}
|
Javascript
let N = 3
function maxPathSum(tri, n, dp)
{
for(let j = 0; j < n; j++ ){
dp[n-1][j] = tri[n-1][j] ;
}
for(let i = n-2; i >= 0; i--){
for(let j = i; j >= 0; j-- ){
dp[i][j] = tri[i][j] + Math.max(dp[i+1][j] , dp[i+1][j+1]) ;
}
}
return dp[0][0] ;
}
let tri = [ [1, 0, 0],
[4, 8, 0],
[1, 5, 3] ];
let dp = new Array(N);
for(let i = 0; i < N; i++)
dp[i] = new Array(N).fill(-1);
console.log(maxPathSum(tri, N, dp));
|
Complexity Analysis:
- Time Complexity: O(m*n) where m = no of rows and n = no of columns
- Auxiliary Space: O(n2)
Method 4: Space Optimization (Without Changning input matrix)
We do not need a 2d matrix we only need a 1d array that stores the minimum of the immediate next column and thus we can reduce space
C++
#include<bits/stdc++.h>
using namespace std;
#define N 3
int maxPathSum(int tri[][N], int n, vector<vector<int>> &dp)
{
vector<int> front(n, -1) , curr(n, -1) ;
for(int j = 0; j < n; j++ ){
front[j] = tri[n-1][j] ;
}
for(int i = n-2; i >= 0; i--){
for(int j = i; j >= 0; j-- ){
curr[j] = tri[i][j] + max(front[j] , front[j+1]) ;
}
front = curr ;
}
return front[0] ;
}
int main()
{
int tri[N][N] = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
vector<vector<int>> dp(N, vector<int>(N, -1) ) ;
cout << maxPathSum(tri, N, dp);
return 0;
}
|
Java
public class GFG
{
static int maxPathSum(int[][] tri, int n, int[][] dp)
{
int front[] = new int[n];
int curr[] = new int[n];
for (int j = 0; j < n; j++) {
front[j] = tri[n - 1][j];
curr[j] = -1;
}
for (int i = n - 2; i >= 0; i--) {
for (int j = i; j >= 0; j--) {
curr[j]
= tri[i][j]
+ Math.max(front[j], front[j + 1]);
}
front = curr;
}
return front[0];
}
public static void main(String[] args)
{
int N = 3;
int[][] tri
= { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
int[][] dp = new int[N][N];
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
dp[i][j] = -1;
}
}
System.out.println(maxPathSum(tri, N, dp));
}
}
|
Python3
N=3
def maxPathSum(tri,n,dp):
front=[-1]*n
curr=[-1]*n
for j in range(n):
front[j] = tri[n-1][j]
for i in range(n-2,-1,-1):
for j in range(i,-1,-1):
curr[j] = tri[i][j] + max(front[j] ,front[j+1])
front=curr
return front[0]
tri=[[1, 0, 0],[4,8,0],[1,5,3]]
dp= [[-1 for i in range(N)] for j in range(N)]
print(maxPathSum(tri,N,dp))
|
C#
using System;
public class GFG {
static int maxPathSum(int[, ] tri, int n, int[, ] dp)
{
int[] front = new int[n];
int[] curr = new int[n];
for (int j = 0; j < n; j++) {
front[j] = tri[n - 1, j];
curr[j] = -1;
}
for (int i = n - 2; i >= 0; i--) {
for (int j = i; j >= 0; j--) {
curr[j]
= tri[i, j]
+ Math.Max(front[j], front[j + 1]);
}
front = curr;
}
return front[0];
}
public static void Main(string[] args)
{
int N = 3;
int[, ] tri
= { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
int[, ] dp = new int[N, N];
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
dp[i, j] = -1;
}
}
Console.WriteLine(maxPathSum(tri, N, dp));
}
}
|
Javascript
let N = 3
function maxPathSum(tri, n, dp)
{
let front= new Array(n).fill(-1);
let curr=new Array(n).fill(-1);
for(let j = 0; j < n; j++ ){
front[j] = tri[n-1][j] ;
}
for(let i = n-2; i >= 0; i--){
for(let j = i; j >= 0; j-- ){
curr[j] = tri[i][j] + Math.max(front[j] , front[j+1]) ;
}
front = curr ;
}
return front[0] ;
}
let tri = [ [1, 0, 0],[4, 8, 0],[1, 5, 3]];
dp = new Array(N);
for(let i = 0; i < N; i++)
dp[i] = new Array(N).fill(-1);
console.log(maxPathSum(tri, N, dp));
|
Complexity Analysis:
- Time Complexity: O(m*n) where m = no of rows and n = no of columns
- Auxiliary Space: O(n)
Method 5: Space Optimization (Changing input matrix)
Applying, DP in bottom-up manner we should solve our problem as:
Example:
3
7 4
2 4 6
8 5 9 3
Step 1 :
3 0 0 0
7 4 0 0
2 4 6 0
8 5 9 3
Step 2 :
3 0 0 0
7 4 0 0
10 13 15 0
Step 3 :
3 0 0 0
20 19 0 0
Step 4:
23 0 0 0
output : 23
C++
#include<bits/stdc++.h>
using namespace std;
#define N 3
int maxPathSum(int tri[][N], int m, int n)
{
for (int i=m-1; i>=0; i--)
{
for (int j=0; j<=i; j++)
{
if (tri[i+1][j] > tri[i+1][j+1])
tri[i][j] += tri[i+1][j];
else
tri[i][j] += tri[i+1][j+1];
}
}
return tri[0][0];
}
int main()
{
int tri[N][N] = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
cout << maxPathSum(tri, 2, 2);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int N = 3;
static int maxPathSum(int tri[][], int m, int n)
{
for (int i = m - 1; i >= 0; i--)
{
for (int j = 0; j <= i; j++)
{
if (tri[i + 1][j] > tri[i + 1][j + 1])
tri[i][j] += tri[i + 1][j];
else
tri[i][j] += tri[i + 1][j + 1];
}
}
return tri[0][0];
}
public static void main (String[] args)
{
int tri[][] = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
System.out.println ( maxPathSum(tri, 2, 2));
}
}
|
Python3
N = 3
def maxPathSum(tri, m, n):
for i in range(m-1, -1, -1):
for j in range(i+1):
if (tri[i+1][j] > tri[i+1][j+1]):
tri[i][j] += tri[i+1][j]
else:
tri[i][j] += tri[i+1][j+1]
return tri[0][0]
tri = [[1, 0, 0],
[4, 8, 0],
[1, 5, 3]]
print(maxPathSum(tri, 2, 2))
|
C#
using System;
class GFG {
static int maxPathSum(int [,]tri,
int m, int n)
{
for (int i = m - 1; i >= 0; i--)
{
for (int j = 0; j <= i; j++)
{
if (tri[i + 1,j] >
tri[i + 1,j + 1])
tri[i,j] +=
tri[i + 1,j];
else
tri[i,j] +=
tri[i + 1,j + 1];
}
}
return tri[0,0];
}
public static void Main ()
{
int [,]tri = { {1, 0, 0},
{4, 8, 0},
{1, 5, 3} };
Console.Write (
maxPathSum(tri, 2, 2));
}
}
|
PHP
<?php
function maxPathSum($tri, $m, $n)
{
for ( $i = $m - 1; $i >= 0; $i--)
{
for ($j = 0; $j <= $i; $j++)
{
if ($tri[$i + 1][$j] > $tri[$i + 1]
[$j + 1])
$tri[$i][$j] += $tri[$i + 1][$j];
else
$tri[$i][$j] += $tri[$i + 1]
[$j + 1];
}
}
return $tri[0][0];
}
$tri= array(array(1, 0, 0),
array(4, 8, 0),
array(1, 5, 3));
echo maxPathSum($tri, 2, 2);
?>
|
Javascript
<script>
let N = 3;
function maxPathSum(tri, m, n)
{
for (let i = m - 1; i >= 0; i--)
{
for (let j = 0; j <= i; j++)
{
if (tri[i + 1][j] > tri[i + 1][j + 1])
tri[i][j] += tri[i + 1][j];
else
tri[i][j] += tri[i + 1][j + 1];
}
}
return tri[0][0];
}
let tri = [[1, 0, 0],
[4, 8, 0],
[1, 5, 3]];
document.write( maxPathSum(tri, 2, 2));
</script>
|
Complexity Analysis:
- Time Complexity: O(m*n) where m = no of rows and n = no of columns
- Auxiliary Space: O(1)
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Find the maximum sum path in the given Matrix
Consider an infinite matrix. The cells of the matrix are filled with natural numbers from the cell (1, 1) in the direction of top right to bottom left diagonally. Then the task is to output the Maximum path sum between two points (X1, Y1) and (X2, Y2). Matrix filling process is as follows: Examples:
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