Maximum triplet sum in array
Last Updated :
05 Nov, 2021
Given an array, the task is to find the maximum triplet sum in the array.
Examples :
Input : arr[] = {1, 2, 3, 0, -1, 8, 10}
Output : 21
10 + 8 + 3 = 21
Input : arr[] = {9, 8, 20, 3, 4, -1, 0}
Output : 37
20 + 9 + 8 = 37
Naive approach: In this method, we simply run three-loop and one by one add three-element and compare with the previous sum if the sum of three-element is greater than store in the previous sum.
C++
#include <bits/stdc++.h>
using namespace std;
int maxTripletSum( int arr[], int n)
{
int sum = INT_MIN;
for ( int i = 0; i < n; i++)
for ( int j = i + 1; j < n; j++)
for ( int k = j + 1; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
int main()
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxTripletSum(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int maxTripletSum( int arr[], int n)
{
int sum = - 1000000 ;
for ( int i = 0 ; i < n; i++)
for ( int j = i + 1 ; j < n; j++)
for ( int k = j + 1 ; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
public static void main(String args[])
{
int arr[] = { 1 , 0 , 8 , 6 , 4 , 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
}
|
Python3
def maxTripletSum(arr, n) :
sm = - 1000000
for i in range ( 0 , n) :
for j in range (i + 1 , n) :
for k in range (j + 1 , n) :
if (sm < (arr[i] + arr[j] + arr[k])) :
sm = arr[i] + arr[j] + arr[k]
return sm
arr = [ 1 , 0 , 8 , 6 , 4 , 2 ]
n = len (arr)
print (maxTripletSum(arr, n))
|
C#
using System;
class GFG {
static int maxTripletSum( int [] arr, int n)
{
int sum = -1000000;
for ( int i = 0; i < n; i++)
for ( int j = i + 1; j < n; j++)
for ( int k = j + 1; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
public static void Main()
{
int [] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
}
|
PHP
<?php
function maxTripletSum( $arr , $n )
{
$sum = PHP_INT_MIN;
for ( $i = 0; $i < $n ; $i ++)
for ( $j = $i + 1; $j < $n ; $j ++)
for ( $k = $j + 1; $k < $n ; $k ++)
if ( $sum < $arr [ $i ] +
$arr [ $j ] +
$arr [ $k ])
$sum = $arr [ $i ] +
$arr [ $j ] +
$arr [ $k ];
return $sum ;
}
$arr = array (1, 0, 8, 6, 4, 2);
$n = count ( $arr );
echo maxTripletSum( $arr , $n );
?>
|
Javascript
<script>
function maxTripletSum(arr, n)
{
let sum = -1000000;
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
for (let k = j + 1; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
let arr = [ 1, 0, 8, 6, 4, 2 ];
let n = arr.length;
document.write(maxTripletSum(arr, n));
</script>
|
Output:
18
Time complexity : O(n^3)
Space complexity : O(1)
Another approach: In this, we first need to sort the whole array and after that when we add the last three-element of the array then we find the maximum sum of triplets.
C++
#include <bits/stdc++.h>
using namespace std;
int maxTripletSum( int arr[], int n)
{
sort(arr, arr + n);
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
int main()
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxTripletSum(arr, n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int maxTripletSum( int arr[], int n)
{
Arrays.sort(arr);
return arr[n - 1 ] + arr[n - 2 ] + arr[n - 3 ];
}
public static void main(String args[])
{
int arr[] = { 1 , 0 , 8 , 6 , 4 , 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
}
|
Python3
def maxTripletSum(arr, n) :
arr.sort()
return (arr[n - 1 ] + arr[n - 2 ] + arr[n - 3 ])
arr = [ 1 , 0 , 8 , 6 , 4 , 2 ]
n = len (arr)
print (maxTripletSum(arr, n))
|
C#
using System;
class GFG {
static int maxTripletSum( int [] arr, int n)
{
Array.Sort(arr);
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
public static void Main()
{
int [] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
}
|
PHP
<?php
function maxTripletSum( $arr , $n )
{
sort( $arr );
return $arr [ $n - 1] + $arr [ $n - 2] +
$arr [ $n - 3];
}
$arr = array ( 1, 0, 8, 6, 4, 2 );
$n = count ( $arr );
echo maxTripletSum( $arr , $n );
?>
|
Javascript
<script>
function maxTripletSum(arr, n)
{
arr.sort();
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
let arr = [ 1, 0, 8, 6, 4, 2 ];
let n = arr.length;
document.write(maxTripletSum(arr, n));
</script>
|
Output:
18
Time complexity: O(nlogn)
Space complexity: O(1)
Efficient approach: Scan the array and compute the Maximum, second maximum, and third maximum element present in the array and return the sum of its and it would be maximum sum.
C++
#include <bits/stdc++.h>
using namespace std;
int maxTripletSum( int arr[], int n)
{
int maxA = INT_MIN, maxB = INT_MIN, maxC = INT_MIN;
for ( int i = 0; i < n; i++) {
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
int main()
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxTripletSum(arr, n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int maxTripletSum( int arr[], int n)
{
int maxA = - 100000000 , maxB = - 100000000 ;
int maxC = - 100000000 ;
for ( int i = 0 ; i < n; i++) {
if (arr[i] > maxA)
{
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
else if (arr[i] > maxB)
{
maxC = maxB;
maxB = arr[i];
}
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
public static void main(String args[])
{
int arr[] = { 1 , 0 , 8 , 6 , 4 , 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
}
|
Python3
def maxTripletSum(arr, n) :
maxA = - 100000000
maxB = - 100000000
maxC = - 100000000
for i in range ( 0 , n) :
if (arr[i] > maxA) :
maxC = maxB
maxB = maxA
maxA = arr[i]
elif (arr[i] > maxB) :
maxC = maxB
maxB = arr[i]
elif (arr[i] > maxC) :
maxC = arr[i]
return (maxA + maxB + maxC)
arr = [ 1 , 0 , 8 , 6 , 4 , 2 ]
n = len (arr)
print (maxTripletSum(arr, n))
|
C#
using System;
class GFG {
static int maxTripletSum( int [] arr, int n)
{
int maxA = -100000000, maxB = -100000000;
int maxC = -100000000;
for ( int i = 0; i < n; i++) {
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
public static void Main()
{
int [] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
}
|
PHP
<?php
function maxTripletSum( $arr , $n )
{
$maxA = PHP_INT_MIN;
$maxB = PHP_INT_MIN;
$maxC = PHP_INT_MIN;
for ( $i = 0; $i < $n ; $i ++)
{
if ( $arr [ $i ] > $maxA )
{
$maxC = $maxB ;
$maxB = $maxA ;
$maxA = $arr [ $i ];
}
else if ( $arr [ $i ] > $maxB )
{
$maxC = $maxB ;
$maxB = $arr [ $i ];
}
else if ( $arr [ $i ] > $maxC )
$maxC = $arr [ $i ];
}
return ( $maxA + $maxB + $maxC );
}
$arr = array ( 1, 0, 8, 6, 4, 2 );
$n = count ( $arr );
echo maxTripletSum( $arr , $n );
?>
|
Javascript
<script>
function maxTripletSum(arr, n)
{
let maxA = Number.MIN_SAFE_INTEGER;
let maxB = Number.MIN_SAFE_INTEGER;
let maxC = Number.MIN_SAFE_INTEGER;
for (let i = 0; i < n; i++) {
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
let arr = [ 1, 0, 8, 6, 4, 2 ];
let n = arr.length;
document.write(maxTripletSum(arr, n));
</script>
|
Output:
18
Time complexity : O(n)
Space complexity : O(1)
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