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Maximum product of a triplet (subsequence of size 3) in array in C++ Program.
In this problem, we are given an array arr[] consisting of n integers. Our task is to find the maximum product of a triplet (subsequence of size 3) in array. Here, we will be finding the triple with maximum product value and then return the product.
Let’s take an example to understand the problem,
Input
arr[] = {9, 5, 2, 11, 7, 4}Output
693
Explanation
Here, we will find the triplet that gives the maximum product of all elements of the array. maxProd = 9 * 11 * 7 = 693
Solution Approach
There can be multiple solutions to the problem. We will be discussing them here,
Method 1
Direct method In this method, we will directly loop through the array and then find all possible triplet. Find the product of elements of each triplet and return the maximum of all of them.
Algorithm
Initialise
maxProd = −1000
Step 1:
Create three nested loops: Loop 1:i −> 0 to n−3 Loop 2: j −> i to n−2 Loop 3: k −> j to n−1
Step 1.1 −
Find the product, prod = arr[i]*arr[j]*arr[k].
Step 1.2 −
if prod > maxProd −> maxProd = prod.
Step 3 −
return maxProd.
Example
Program to show the implementation of our solution,
#include <iostream>
using namespace std;
int calcMaxProd(int arr[], int n){
int maxProd = −1000;
int prod;
for (int i = 0; i < n − 2; i++)
for (int j = i + 1; j < n − 1; j++)
for (int k = j + 1; k < n; k++){
prod = arr[i] * arr[j] * arr[k];
if(maxProd < prod)
maxProd = prod;
}
return maxProd;
}
int main(){
int arr[] = { 9, 5, 2, 11, 7, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"Maximum product of a triplet in array is "<<calcMaxProd(arr, n);
return 0;
}Output
Maximum product of a triplet in array is 693
Method 2
Using Sorting
In this method, we will sort the array in descending order. In sorted array, the maximum product triplet will be at,
(arr[0], arr[1], arr[2]) (arr[0], arr[1], arr[2])
We will return the maximum of the product of these triplets.
Algorithm
Step 1 −
Sort the given array in descending order.
Step 2 −
Find product of triples, maxTriplet1 = arr[0]*arr[1]*arr[2] maxTriplet2 = arr[0]*arr[n−1]*arr[n−2]
Step 3 −
if( maxTriplet1 > maxTriplet2 ) −> return maxTriplet1
Step 4 −
else −> return maxTriplet2.
Example
Program to illustrate the working of our solution,
#include <bits/stdc++.h>
using namespace std;
int calcMaxProd(int arr[], int n){
sort(arr, arr + n, greater<>());
int maxTriplet1 = arr[0]*arr[1]*arr[2];
int maxTriplet2 = arr[0]*arr[n−1]*arr[n−2];
if(maxTriplet1 > maxTriplet2)
return maxTriplet1;
return maxTriplet2;
}
int main(){
int arr[] = { 9, 5, 2, 11, 7, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"Maximum product of a triplet in array is
"<<calcMaxProd(arr, n);
return 0;
}Output
Maximum product of a triplet in array is 693
Method 3
Finding triplet values.
As we now know that the maximum product triplet can be from either of the two triplets,
(maximum, second_max, third_max) (maximum, minimum, second_min)
So, we can directly find these values by traversing the array, and then using the values, we will find the maximum product triplet.
Algorithm
Initialise
max = -1000, secMax = -1000, thirdMax = -1000 , min = 10000, secMin = 10000
Step 1−
loop the array i −> 0 to n−1.
Step 1.1
if(arr[i] > max) −> thirdMax = secMax, secMax = max, max = arr[i]
Step 1.2−
elseif(arr[i] > secMax) −> thirdMax = secMax, secMax = arr[i]
Step 1.3−
elseif(arr[i] > thirdMax) −> thirdMax = arr[i]
Step 1.4−
if(arr[i] < min) −> secMin = min, min = arr[i]
Step 1.4−
elseif(arr[i] < secMin) −> secMin = arr[i]
Step 2−
triplet1 = max * secMax * thridMax triplet2 = max * min * secMin
Step 3−
if(triplet1 > triplet2) −> return triplet1
Step 4−
else −> return triplet2
Example
Program to illustrate the working of our solution,
#include <iostream>
using namespace std;
int calcMaxProd(int arr[], int n){
int max = −1000, secMax = −1000, thirdMax = −1000;
int min = 1000, secMin = 1000;
for (int i = 0; i < n; i++){
if (arr[i] > max){
thirdMax = secMax;
secMax = max;
max = arr[i];
}
else if (arr[i] > secMax){
thirdMax = secMax;
secMax = arr[i];
}
else if (arr[i] > thirdMax)
thirdMax = arr[i];
if (arr[i] < min){
secMin = min;
min = arr[i];
}
else if(arr[i] < secMin)
secMin = arr[i];
}
int triplet1 = max * secMax * thirdMax;
int triplet2 = max * secMin * min;
if(triplet1 > triplet2)
return triplet1;
return triplet2;
}
int main(){
int arr[] = { 9, 5, 2, 11, 7, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"Maximum product of a triplet in array is
"<<calcMaxProd(arr, n);
return 0;
}Output
Maximum product of a triplet in array is 693
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