Maximum product of sum of two contiguous subarrays of an array
Given an array arr[] of N positive integers, the task is to split the array into two contiguous subarrays such that the product of the sum of two contiguous subarrays is maximum.
Examples:
Input: arr[] = {4, 10, 1, 7, 2, 9}
Output: 270
All possible partitions and their product of sum are:
{4} and {10, 1, 7, 2, 9} -> product of sum = 116
{4, 10} and {1, 7, 2, 9} -> product of sum = 266
{4, 10, 1} and {7, 2, 9} -> product of sum = 270
{4, 10, 1, 7} and {2, 9} -> product of sum = 242
{4, 10, 1, 7, 2} and {9} -> product of sum = 216Input: arr[] = {4, 10, 11, 10, 4}
Output: 350
Naive approach: A simple approach is to consider all the possible partitions for the subarrays one by one and calculate the maximum product of sum for subarrays.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the maximum // product of sum for any partition int maxProdSum(int arr[], int n) { int leftArraySum = 0, maxProduct = 0; // Traversing the array for (int i = 0; i < n; i++) { // Compute left array sum leftArraySum += arr[i]; // Compute right array sum int rightArraySum = 0; for (int j = i + 1; j < n; j++) { rightArraySum += arr[j]; } // Multiplying left and right subarray sum int k = leftArraySum * rightArraySum; // Checking for the maximum product // of sum of left and right subarray if (k > maxProduct) { maxProduct = k; } } // Printing the maximum product return maxProduct; } // Driver code int main() { int arr[] = { 4, 10, 1, 7, 2, 9 }; int n = sizeof(arr) / sizeof(int); cout << maxProdSum(arr, n); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the maximum // product of sum for any partition static int maxProdSum(int arr[], int n) { int leftArraySum = 0, maxProduct = 0; // Traversing the array for (int i = 0; i < n; i++) { // Compute left array sum leftArraySum += arr[i]; // Compute right array sum int rightArraySum = 0; for (int j = i + 1; j < n; j++) { rightArraySum += arr[j]; } // Multiplying left and right subarray sum int k = leftArraySum * rightArraySum; // Checking for the maximum product // of sum of left and right subarray if (k > maxProduct) { maxProduct = k; } } // Printing the maximum product return maxProduct; } // Driver code public static void main(String[] args) { int arr[] = { 4, 10, 1, 7, 2, 9 }; int n = arr.length; System.out.print(maxProdSum(arr, n)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach # Function to return the maximum # product of sum for any partition def maxProdSum(arr, n): leftArraySum = 0; maxProduct = 0; # Traversing the array for i in range(n): # Compute left array sum leftArraySum += arr[i]; # Compute right array sum rightArraySum = 0; for j in range(i + 1, n): rightArraySum += arr[j]; # Multiplying left and right subarray sum k = leftArraySum * rightArraySum; # Checking for the maximum product # of sum of left and right subarray if (k > maxProduct): maxProduct = k; # Printing the maximum product return maxProduct; # Driver code if __name__ == '__main__': arr = [ 4, 10, 1, 7, 2, 9 ]; n = len(arr); print(maxProdSum(arr, n)); # This code is contributed by Rajput-Ji |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the maximum // product of sum for any partition static int maxProdSum(int []arr, int n) { int leftArraySum = 0, maxProduct = 0; // Traversing the array for (int i = 0; i < n; i++) { // Compute left array sum leftArraySum += arr[i]; // Compute right array sum int rightArraySum = 0; for (int j = i + 1; j < n; j++) { rightArraySum += arr[j]; } // Multiplying left and right subarray sum int k = leftArraySum * rightArraySum; // Checking for the maximum product // of sum of left and right subarray if (k > maxProduct) { maxProduct = k; } } // Printing the maximum product return maxProduct; } // Driver code public static void Main(String[] args) { int []arr = { 4, 10, 1, 7, 2, 9 }; int n = arr.Length; Console.Write(maxProdSum(arr, n)); } } // This code is contributed by PrinciRaj1992 |
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Output:
270
Time Complexity: O(N2)
Efficient approach: A better approach is to use the concept of prefix array sum which helps in calculating the sum of both the contiguous subarrays.
- Prefix sum of the elements can be calculated.
- The left array sum is the value at ith position.
- The right array sum is the value at the last position – ith position.
- Now calculate k, the product of sum of these left and right subarrays.
- Find and print the maximum of the product of these arrays.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the maximum // product of sum for any partition int maxProdSum(int arr[], int n) { int prefixArraySum[n], maxProduct = 0; // Initialise prefixArraySum[0] // with arr[0] element prefixArraySum[0] = arr[0]; // Traverse array elements // to compute prefix array sum for (int i = 1; i < n; i++) { prefixArraySum[i] = prefixArraySum[i - 1] + arr[i]; } for (int i = 0; i < n - 1; i++) { // Compute left and right array sum int leftArraySum = prefixArraySum[i]; int rightArraySum = prefixArraySum[n - 1] - prefixArraySum[i]; // Multiplying left and right subarray sum int k = leftArraySum * rightArraySum; // Checking for maximum product of // the sum of left and right subarray if (k > maxProduct) { maxProduct = k; } } // Printing the maximum value return maxProduct; } // Driver code int main() { int arr[] = { 4, 10, 1, 7, 2, 9 }; int n = sizeof(arr) / sizeof(int); cout << maxProdSum(arr, n); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the maximum // product of sum for any partition static int maxProdSum(int arr[], int n) { int []prefixArraySum = new int[n]; int maxProduct = 0; // Initialise prefixArraySum[0] // with arr[0] element prefixArraySum[0] = arr[0]; // Traverse array elements // to compute prefix array sum for (int i = 1; i < n; i++) { prefixArraySum[i] = prefixArraySum[i - 1] + arr[i]; } for (int i = 0; i < n - 1; i++) { // Compute left and right array sum int leftArraySum = prefixArraySum[i]; int rightArraySum = prefixArraySum[n - 1] - prefixArraySum[i]; // Multiplying left and right subarray sum int k = leftArraySum * rightArraySum; // Checking for maximum product of // the sum of left and right subarray if (k > maxProduct) { maxProduct = k; } } // Printing the maximum value return maxProduct; } // Driver code public static void main(String[] args) { int arr[] = { 4, 10, 1, 7, 2, 9 }; int n = arr.length; System.out.print(maxProdSum(arr, n)); } } // This code is contributed by PrinciRaj1992 |
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Python3
# Python implementation of the approach # Function to return the maximum # product of sum for any partition def maxProdSum(arr, n): prefixArraySum = [0] * n maxProduct = 0 # Initialise prefixArraySum[0] # with arr[0] element prefixArraySum[0] = arr[0] # Traverse array elements # to compute prefix array sum for i in range(1, n): prefixArraySum[i] = prefixArraySum[i - 1] + arr[i] for i in range(n - 1): # Compute left and right array sum leftArraySum = prefixArraySum[i] rightArraySum = prefixArraySum[n - 1] - \ prefixArraySum[i] # Multiplying left and right subarray sum k = leftArraySum * rightArraySum # Checking for maximum product of # the sum of left and right subarray if (k > maxProduct): maxProduct = k # Printing the maximum value return maxProduct # Driver code arr = [4, 10, 1, 7, 2, 9] n = len(arr) print(maxProdSum(arr, n)) # This code is contributed by SHUBHAMSINGH10 |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the maximum // product of sum for any partition static int maxProdSum(int []arr, int n) { int []prefixArraySum = new int[n]; int maxProduct = 0; // Initialise prefixArraySum[0] // with arr[0] element prefixArraySum[0] = arr[0]; // Traverse array elements // to compute prefix array sum for (int i = 1; i < n; i++) { prefixArraySum[i] = prefixArraySum[i - 1] + arr[i]; } for (int i = 0; i < n - 1; i++) { // Compute left and right array sum int leftArraySum = prefixArraySum[i]; int rightArraySum = prefixArraySum[n - 1] - prefixArraySum[i]; // Multiplying left and right subarray sum int k = leftArraySum * rightArraySum; // Checking for maximum product of // the sum of left and right subarray if (k > maxProduct) { maxProduct = k; } } // Printing the maximum value return maxProduct; } // Driver code public static void Main(String[] args) { int []arr = { 4, 10, 1, 7, 2, 9 }; int n = arr.Length; Console.Write(maxProdSum(arr, n)); } } // This code is contributed by Rajput-Ji |
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Output:
270
Time Complexity: O(n)
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