# Count of ways to split an Array into three contiguous Subarrays having increasing Sum

Given an array arr[] consisting of non-negative integers, the task is to find the number of ways to split the array into three non-empty contiguous subarrays such that their respective sum of elements are in increasing order.

Examples:

Input: arr[] = {2, 3, 1, 7}
Output:
Explanation:
{{2}, {3, 1}, {7}}, {{2}, {3}, {1, 7}} are the possible splits.

Input: arr[] = {1, 2, 0}
Output: 0

Approach: The idea is to use the Prefix and Suffix sum array and Two Pointers technique. Follow the steps below to solve the problem:

• Generate the prefix sum array and suffix sum array.
• Initialize two pointers s and e to find the sum of the second subarray.
• Iterate over the array, increment curr_subarray_sum with arr[e] while curr_subarray_sum less than prefix_sum[s – 1] and keep incrementing e.
• Whenever curr_subarray_sum is ? prefix_sum[s – 1], then check if curr_subarray_sum is ? suffix_sum[e]. If found to be true, increment count.
• Reduce curr_subarray_sum by arr[s] and increment s.
• Repeat the above steps and finally, print count

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach` `#include` `using` `namespace` `std;`   `// Function to count the number of ways` `// to split array into three contiguous` `// subarrays of the required type` `int` `findCount(``int` `arr[], ``int` `n)` `{`   `    ``// Stores the prefix sums` `    ``int` `prefix_sum[n];`   `    ``prefix_sum[0] = arr[0];`   `    ``for``(``int` `i = 1; i < n; i++)` `        ``prefix_sum[i] = prefix_sum[i - 1] + arr[i];`   `    ``// Stores the suffix sums` `    ``int` `suffix_sum[n];`   `    ``suffix_sum[n - 1] = arr[n - 1];`   `    ``for``(``int` `i = n - 2; i >= 0; i--)` `        ``suffix_sum[i] = suffix_sum[i + 1] + arr[i];`   `    ``int` `s = 1, e = 1;` `    ``int` `curr_subarray_sum = 0, count = 0;`   `    ``// Traverse the given array` `    ``while` `(s < n - 1 && e < n - 1)` `    ``{` `        `  `        ``// Updating curr_subarray_sum until` `        ``// it is less than prefix_sum[s-1]` `        ``while` `(e < n - 1 && curr_subarray_sum < ` `               ``prefix_sum[s - 1]) ` `        ``{` `            ``curr_subarray_sum += arr[e++];` `        ``}`   `        ``if` `(curr_subarray_sum <= suffix_sum[e])` `        ``{` `            `  `            ``// Increase count` `            ``count++;` `        ``}`   `        ``// Decrease curr_subarray_sum by arr[s[]` `        ``curr_subarray_sum -= arr[s++];` `    ``}`   `    ``// Return count` `    ``return` `count;` `}`   `// Driver code` `int32_t main()` `{` `    ``int` `arr[] = { 2, 3, 1, 7 };` `    ``int` `n = ``sizeof` `arr / ``sizeof` `arr[0];` `    `  `    ``cout << (findCount(arr, n));` `}`   `// This code is contributed by Stream_Cipher`

## Java

 `// Java Program to implement` `// the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to count the number of ways` `    ``// to split array into three contiguous` `    ``// subarrays of the required type` `    ``static` `int` `findCount(``int` `arr[], ``int` `n)` `    ``{`   `        ``// Stores the prefix sums` `        ``int``[] prefix_sum = ``new` `int``[n];`   `        ``prefix_sum[``0``] = arr[``0``];`   `        ``for` `(``int` `i = ``1``; i < n; i++)` `            ``prefix_sum[i]` `                ``= prefix_sum[i - ``1``] + arr[i];`   `        ``// Stores the suffix sums` `        ``int``[] suffix_sum = ``new` `int``[n];`   `        ``suffix_sum[n - ``1``] = arr[n - ``1``];`   `        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--)` `            ``suffix_sum[i]` `                ``= suffix_sum[i + ``1``] + arr[i];`   `        ``int` `s = ``1``, e = ``1``;` `        ``int` `curr_subarray_sum = ``0``, count = ``0``;`   `        ``// Traverse the given array` `        ``while` `(s < n - ``1` `&& e < n - ``1``) {`   `            ``// Updating curr_subarray_sum until` `            ``// it is less than prefix_sum[s-1]` `            ``while` `(e < n - ``1` `                   ``&& curr_subarray_sum` `                          ``< prefix_sum[s - ``1``]) {` `                ``curr_subarray_sum += arr[e++];` `            ``}`   `            ``if` `(curr_subarray_sum <= suffix_sum[e]) {` `                ``// Increase count` `                ``count++;` `            ``}`   `            ``// Decrease curr_subarray_sum by arr[s[]` `            ``curr_subarray_sum -= arr[s++];` `        ``}`   `        ``// Return count` `        ``return` `count;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String args[])` `    ``{`   `        ``int``[] arr = { ``2``, ``3``, ``1``, ``7` `};` `        ``int` `n = arr.length;` `        ``System.out.println(findCount(arr, n));` `    ``}` `}`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to count the number of ways` `# to split array into three contiguous` `# subarrays of the required type` `def` `findCount(arr, n):`   `    ``# Stores the prefix sums` `    ``prefix_sum ``=` `[``0` `for` `x ``in` `range``(n)]` `    ``prefix_sum[``0``] ``=` `arr[``0``]` `    `  `    ``for` `i ``in` `range``(``1``, n):` `        ``prefix_sum[i] ``=` `prefix_sum[i ``-` `1``] ``+` `arr[i]` `    `  `    ``# Stores the suffix sums` `    ``suffix_sum ``=` `[``0` `for` `x ``in` `range``(n)]` `    `  `    ``suffix_sum[n ``-` `1``] ``=` `arr[n ``-` `1``]` `    `  `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``):` `        ``suffix_sum[i] ``=` `suffix_sum[i ``+` `1``] ``+` `arr[i]` `    `  `    ``s ``=` `1` `    ``e ``=` `1` `    ``curr_subarray_sum ``=` `0` `    ``count ``=` `0` `    `  `    ``#Traverse the given array` `    ``while` `(s < n ``-` `1` `and` `e < n ``-` `1``):` `        `  `        ``# Updating curr_subarray_sum until` `        ``# it is less than prefix_sum[s-1]` `        ``while` `(e < n ``-` `1` `and` `               ``curr_subarray_sum < prefix_sum[s ``-` `1``]):` `            ``curr_subarray_sum ``+``=` `arr[e]` `            ``e ``+``=` `1` `                `  `        ``if` `(curr_subarray_sum <``=` `suffix_sum[e]):` `            `  `            ``# Increase count` `            ``count ``+``=` `1` `                `  `        ``# Decrease curr_subarray_sum by arr[s[]` `        ``curr_subarray_sum ``-``=` `arr[s]` `        ``s ``+``=` `1` `    `  `    ``# Return count` `    ``return` `count` `    `  `# Driver code` `arr ``=` `[ ``2``, ``3``, ``1``, ``7` `]` `n ``=` `len``(arr) `   `print``(findCount(arr, n))`   `# This code is contributed by Stream_Cipher`

## C#

 `// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{`   `  ``// Function to count the number of ways` `  ``// to split array into three contiguous` `  ``// subarrays of the required type` `  ``static` `int` `findCount(``int` `[]arr, ``int` `n)` `  ``{`   `    ``// Stores the prefix sums` `    ``int``[] prefix_sum = ``new` `int``[n];`   `    ``prefix_sum[0] = arr[0];`   `    ``for` `(``int` `i = 1; i < n; i++)` `      ``prefix_sum[i] = prefix_sum[i - 1] + arr[i];`   `    ``// Stores the suffix sums` `    ``int``[] suffix_sum = ``new` `int``[n];`   `    ``suffix_sum[n - 1] = arr[n - 1];`   `    ``for` `(``int` `i = n - 2; i >= 0; i--)` `      ``suffix_sum[i] = suffix_sum[i + 1] + arr[i];`   `    ``int` `s = 1, e = 1;` `    ``int` `curr_subarray_sum = 0, count = 0;`   `    ``// Traverse the given array` `    ``while` `(s < n - 1 && e < n - 1) ` `    ``{`   `      ``// Updating curr_subarray_sum until` `      ``// it is less than prefix_sum[s-1]` `      ``while` `(e < n - 1 && ` `             ``curr_subarray_sum < prefix_sum[s - 1]) ` `      ``{` `        ``curr_subarray_sum += arr[e++];` `      ``}`   `      ``if` `(curr_subarray_sum <= suffix_sum[e])` `      ``{` `        ``// Increase count` `        ``count++;` `      ``}`   `      ``// Decrease curr_subarray_sum by arr[s[]` `      ``curr_subarray_sum -= arr[s++];` `    ``}`   `    ``// Return count` `    ``return` `count;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(String []args)` `  ``{`   `    ``int``[] arr = { 2, 3, 1, 7 };` `    ``int` `n = arr.Length;` `    ``Console.WriteLine(findCount(arr, n));` `  ``}` `}`   `// This code is contributed by Rohit_ranjan`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(N)

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