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++ code to find maximum triplet sum #include <bits/stdc++.h> using namespace std;
int maxTripletSum( int arr[], int n)
{ // Initialize sum with INT_MIN
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;
} // Driven code int main()
{ int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxTripletSum(arr, n);
return 0;
} |
// Java code to find maximum triplet sum import java.io.*;
class GFG {
static int maxTripletSum( int arr[], int n)
{
// Initialize sum with INT_MIN
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;
}
// Driven code
public static void main(String args[])
{
int arr[] = { 1 , 0 , 8 , 6 , 4 , 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
} // This code is contributed by Nikita Tiwari. |
# Python 3 code to find # maximum triplet sum def maxTripletSum(arr, n) :
# Initialize sum with
# INT_MIN
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
# Driven code arr = [ 1 , 0 , 8 , 6 , 4 , 2 ]
n = len (arr)
print (maxTripletSum(arr, n))
# This code is contributed by Nikita Tiwari. |
// C# code to find maximum triplet sum using System;
class GFG {
static int maxTripletSum( int [] arr, int n)
{
// Initialize sum with INT_MIN
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;
}
// Driven code
public static void Main()
{
int [] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
} // This code is contributed by vt_m. |
<?php // PHP code to find maximum triplet sum function maxTripletSum( $arr , $n )
{ // Initialize sum with INT_MIN
$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 ;
} // Driver Code
$arr = array (1, 0, 8, 6, 4, 2);
$n = count ( $arr );
echo maxTripletSum( $arr , $n );
// This code is contributed by anuj_67. ?> |
<script> // JavaScript Program to find maximum triplet sum function maxTripletSum(arr, n)
{
// Initialize sum with INT_MIN
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;
}
// Driver code let arr = [ 1, 0, 8, 6, 4, 2 ];
let n = arr.length;
document.write(maxTripletSum(arr, n));
</script> // This code is contributed by sanjoy_62. |
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++ code to find maximum triplet sum #include <bits/stdc++.h> using namespace std;
// This function assumes that there are at least // three elements in arr[]. int maxTripletSum( int arr[], int n)
{ // sort the given array
sort(arr, arr + n);
// After sorting the array.
// Add last three element of the given array
return arr[n - 1] + arr[n - 2] + arr[n - 3];
} // Driven code int main()
{ int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxTripletSum(arr, n);
return 0;
} |
// Java code to find maximum triplet sum import java.io.*;
import java.util.*;
class GFG {
// This function assumes that there are
// at least three elements in arr[].
static int maxTripletSum( int arr[], int n)
{
// sort the given array
Arrays.sort(arr);
// After sorting the array.
// Add last three element
// of the given array
return arr[n - 1 ] + arr[n - 2 ] + arr[n - 3 ];
}
// Driven code
public static void main(String args[])
{
int arr[] = { 1 , 0 , 8 , 6 , 4 , 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
} // This code is contributed by Nikita Tiwari. |
# Python 3 code to find # maximum triplet sum # This function assumes # that there are at least # three elements in arr[]. def maxTripletSum(arr, n) :
# sort the given array
arr.sort()
# After sorting the array.
# Add last three element
# of the given array
return (arr[n - 1 ] + arr[n - 2 ] + arr[n - 3 ])
# Driven code arr = [ 1 , 0 , 8 , 6 , 4 , 2 ]
n = len (arr)
print (maxTripletSum(arr, n))
# This code is contributed by Nikita Tiwari. |
// C# code to find maximum triplet sum using System;
class GFG {
// This function assumes that there are
// at least three elements in arr[].
static int maxTripletSum( int [] arr, int n)
{
// sort the given array
Array.Sort(arr);
// After sorting the array.
// Add last three element
// of the given array
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
// Driven code
public static void Main()
{
int [] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
} // This code is contributed by vt_m. |
<?php // PHP code to find // maximum triplet sum // This function assumes that // there are at least // three elements in arr[]. function maxTripletSum( $arr , $n )
{ // sort the given array
sort( $arr );
// After sorting the array.
// Add last three element
// of the given array
return $arr [ $n - 1] + $arr [ $n - 2] +
$arr [ $n - 3];
} // Driver code $arr = array ( 1, 0, 8, 6, 4, 2 );
$n = count ( $arr );
echo maxTripletSum( $arr , $n );
// This code is contributed by anuj_67. ?> |
<script> //Javascript code to find maximum triplet sum // This function assumes that there are at least // three elements in arr[]. function maxTripletSum(arr, n)
{ // sort the given array
arr.sort();
// After sorting the array.
// Add last three element of the given array
return arr[n - 1] + arr[n - 2] + arr[n - 3];
} // Driven code let arr = [ 1, 0, 8, 6, 4, 2 ];
let n = arr.length;
document.write(maxTripletSum(arr, n));
// This code is contributed by Mayank Tyagi </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++ code to find maximum triplet sum #include <bits/stdc++.h> using namespace std;
// This function assumes that there are at least // three elements in arr[]. int maxTripletSum( int arr[], int n)
{ // Initialize Maximum, second maximum and third
// maximum element
int maxA = INT_MIN, maxB = INT_MIN, maxC = INT_MIN;
for ( int i = 0; i < n; i++) {
// Update Maximum, second maximum and third
// maximum element
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third maximum
// element
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
} // Driven code int main()
{ int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxTripletSum(arr, n);
return 0;
} |
// Java code to find maximum triplet sum import java.io.*;
import java.util.*;
class GFG {
// This function assumes that there
// are at least three elements in arr[].
static int maxTripletSum( int arr[], int n)
{
// Initialize Maximum, second maximum and third
// maximum element
int maxA = - 100000000 , maxB = - 100000000 ;
int maxC = - 100000000 ;
for ( int i = 0 ; i < n; i++) {
// Update Maximum, second maximum
// and third maximum element
if (arr[i] > maxA)
{
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third maximum
// element
else if (arr[i] > maxB)
{
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
// Driven code
public static void main(String args[])
{
int arr[] = { 1 , 0 , 8 , 6 , 4 , 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
} // This code is contributed by Nikita Tiwari. |
# Python 3 code to find # maximum triplet sum # This function assumes # that there are at least # three elements in arr[]. def maxTripletSum(arr, n) :
# Initialize Maximum, second
# maximum and third maximum
# element
maxA = - 100000000
maxB = - 100000000
maxC = - 100000000
for i in range ( 0 , n) :
# Update Maximum, second maximum
# and third maximum element
if (arr[i] > maxA) :
maxC = maxB
maxB = maxA
maxA = arr[i]
# Update second maximum and
# third maximum element
elif (arr[i] > maxB) :
maxC = maxB
maxB = arr[i]
# Update third maximum element
elif (arr[i] > maxC) :
maxC = arr[i]
return (maxA + maxB + maxC)
# Driven code arr = [ 1 , 0 , 8 , 6 , 4 , 2 ]
n = len (arr)
print (maxTripletSum(arr, n))
# This code is contributed by Nikita Tiwari. |
// C# code to find maximum triplet sum using System;
class GFG {
// This function assumes that there
// are at least three elements in arr[].
static int maxTripletSum( int [] arr, int n)
{
// Initialize Maximum, second maximum
// and third maximum element
int maxA = -100000000, maxB = -100000000;
int maxC = -100000000;
for ( int i = 0; i < n; i++) {
// Update Maximum, second maximum
// and third maximum element
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third
// maximum element
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
// Driven code
public static void Main()
{
int [] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
} // This code is contributed by vt_m. |
<?php // PHP code to find // maximum triplet sum // This function assumes that // there are at least three // elements in arr[]. function maxTripletSum( $arr , $n )
{ // Initialize Maximum,
// second maximum and
// third maximum element
$maxA = PHP_INT_MIN;
$maxB = PHP_INT_MIN;
$maxC = PHP_INT_MIN;
for ( $i = 0; $i < $n ; $i ++)
{
// Update Maximum,
// second maximum and
// third maximum element
if ( $arr [ $i ] > $maxA )
{
$maxC = $maxB ;
$maxB = $maxA ;
$maxA = $arr [ $i ];
}
// Update second maximum and
// third maximum element
else if ( $arr [ $i ] > $maxB )
{
$maxC = $maxB ;
$maxB = $arr [ $i ];
}
// Update third maximum element
else if ( $arr [ $i ] > $maxC )
$maxC = $arr [ $i ];
}
return ( $maxA + $maxB + $maxC );
} // Driven code $arr = array ( 1, 0, 8, 6, 4, 2 );
$n = count ( $arr );
echo maxTripletSum( $arr , $n );
// This code is contributed by anuj_67. ?> |
<script> // JavaScript code to find maximum triplet sum // This function assumes that there are at least // three elements in arr[]. function maxTripletSum(arr, n)
{ // Initialize Maximum, second maximum and third
// maximum element
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++) {
// Update Maximum, second maximum and third
// maximum element
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third maximum
// element
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
} // Driven code let arr = [ 1, 0, 8, 6, 4, 2 ];
let n = arr.length;
document.write(maxTripletSum(arr, n));
// This code is contributed by Surbhi Tyagi. </script> |
Output:
18
Time complexity : O(n)
Space complexity : O(1)