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# Solving the Partition Problem Using Kadane's Algorithm

In this chapter of C++ programs, we will learn about:
• kadane's algorithm example
• partition problem c++

Lets begin with kadane's algorithm and partition problem c++.

### Solving the Partition Problem Using Kadane's Algorithm

The Partition Problem is a classic challenge in computer science: can a given set of numbers be divided into two subsets such that the sum of elements in both subsets is equal?

In this tutorial, we'll explore how Kadane's Algorithm, renowned for finding the maximum sum subarray, can be adapted ingeniously to solve this problem efficiently.

### Understanding the Partition Problem C++

The goal of the Partition Problem is to partition an array into two subsets such that their sums are equal. For instance, given an array `[1, 5, 11, 5]`, it can be partitioned into `[1, 5, 5]` and `[11]` with each subset summing to 11.

### Kadane's Algorithm Overview

Kadane's Algorithm is famous for finding the maximum sum subarray in linear time, (O(n)), where n is the number of elements in the array. It works by iterating through the array while maintaining a running maximum sum of subarrays.

#### Adapting Kadane's Algorithm for the Partition Problem

To solve the Partition Problem using Kadane's Algorithm C++, follow these steps:

1. Calculate Total Sum: Compute the total sum S of all elements in the array.

2. Target Sum Calculation: For the array to be partitioned into two subsets with equal sums, each subset should ideally have a sum of S/2, where S is the total sum of the array.

3. Use Kadane’s Algorithm: Modify Kadane's Algorithm to find a subset of elements that sum up to S/2. If such a subset exists, then the remaining elements automatically form the second subset with the same sum.

#### Kadane's algorithm example

Kadane's algorithm example implementation in C++
``````#include <iostream>
#include <vector>
#include <numeric> // for std::accumulate

bool partitionEqualSum(std::vector<int>& arr) {
int totalSum = std::accumulate(arr.begin(), arr.end(), 0);

// If total sum is odd, partition is not possible
if (totalSum % 2 != 0) {
return false;
}

int target = totalSum / 2;
int currentSum = 0;

for (int num : arr) {
currentSum += num;
if (currentSum == target) {
return true;
} else if (currentSum > target) {
return false;
}
}

return false;
}

int main() {
std::vector<int> arr = {1, 5, 11, 5};

if (partitionEqualSum(arr)) {
std::cout << "Partition is possible\n";
} else {
std::cout << "Partition is not possible\n";
}

return 0;
}``````

#### Explanation: kadane's algorithm explained for partition problem

- partitionEqualSum` function: This function takes a reference to a vector of integers (`arr`). It calculates the `totalSum` of all elements in the array using `std::accumulate` from the `<numeric>` header.

- Target Calculation: It checks if the `totalSum` is odd. If it is, partitioning is not possible, so it returns `false`.

- Iterative Check: It iterates through each element in the array, updating `currentSum` and checking if it matches `target`. If `currentSum` equals `target`, it returns `true`, indicating a valid partition. If `currentSum` exceeds `target`, it returns `false`.

- Main Function: In the `main` function, an example vector `arr` is initialized with `{1, 5, 11, 5}`. It calls `partitionEqualSum(arr)` and prints whether partitioning is possible based on the returned boolean value.

This C++ implementation demonstrates how Kadane's Algorithm can be adapted to efficiently solve the Partition Problem by checking if an array can be divided into two subsets with equal sums.

Kadane's Algorithm, originally designed for a different purpose, elegantly solves the Partition Problem by efficiently finding subarrays with specific sum constraints. This adaptation showcases the versatility of algorithmic techniques and underscores the importance of understanding core algorithmic principles in solving complex problems effectively.
This wraps up our session on kadane's algorithm explained for partition problem c++

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