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

• Difficulty Level : Hard
• Last Updated : 08 Sep, 2021

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:

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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 â‰¥ prefix_sum[s – 1], then check if curr_subarray_sum is â‰¤ 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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