Count the number of ways to divide an array into three contiguous parts having equal sum

3.2

Given a array of n numbers. Our task is to find out the number of ways to divide the array into three contiguous parts such that the sum of three parts is equal. In other words, we need to find the number of index pairs i and j such that sum of elements from 0 to i-1 is equal to sum of elements from i to j is equal to the sum of elements from j+1 to n-1.

Examples:

Input  : arr[] = {1, 2, 3, 0, 3}
Output : 2
Following are two possible ways to partition
1) Three parts are (0, 1), (2, 2) and (3, 4)
2) Three parts are (0, 1), (2, 3) and (4, 4)

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

If the sum of all the elements of array is not divisible by 3 return 0. Else it is obvious that the sum of each part of each contiguous part will be equal to sum of all elements divided by 3.

Step 1 : Create an array of same size as given array whose i-th index holds the value of sum of elements from indices 0 to i of the given array. Let’s call it prefix array

Step 2 : Create another array of same size as given array whose i-th index the value of sum of elements from indices i to n-1 of the given array. Let’s call it suffix array.

Step 3 : The idea is simple, we start traversing the prefix array and suppose at the i-th index of the prefix array the value of prefix array is equal to (sum of all elements of given array)/3.

Step 4 : For the i found in the above step we look into the suffix array from (i+2)-th index and wherever the value of suffix array is equal to (sum of all elements of given array)/3, we increase the counter variable by 1.

To implement step 4 we traverse suffix array and wherever the value of suffix array is equal to the sum of all elements of given array/3 we push that index of suffix array into the vector. And we do binary search in the vector to calculate number of values in suffix array which are as according to the step 4.

We search in suffix array because there should be at least 1 element between the first and third part.

For more explanation see implementation below

// C++ program to count number of ways we can
// partition an array in three parts with equal
// sum.
#include<bits/stdc++.h>
using namespace std;

// binary search to find the number of required
// indices in suffix array. Returns index of
// first element which is greater than x.
int binarysearch(vector <int> &v, int x)
{
    int low = 0, high = v.size()-1;
    while (low <= high)
    {
        int mid = (low + high)/2;
        if (v[mid] <= x)
            low = mid + 1;
        else if (v[mid] > x && v[mid-1] <= x)
            return mid;
        else if (v[mid] > x && mid == 0)
            return mid;
        else
            high = mid-1;
    }
    return -1;
}

// function to calculate the number of ways to
// divide the array into three contiguous parts
int CountContiguousParts(int arr[] ,int n)
{
    int count = 0;  // initializing answer

    // Creating and filling  prefix array
    int prefix[n];
    prefix[0] = arr[0];
    for (int i=1; i<n; i++)
        prefix[i] = prefix[i-1] + arr[i];

    // Total sum of elements is equal to last
    // value in prefix array.
    int total_sum = prefix[n-1];

    // If sum of all the elements is not divisible
    // by 3, we can't divide array in 3 parts of
    // same sum.
    if (total_sum%3 != 0)
        return 0;

    // Creating and filling suffix array
    int suffix[n];
    suffix[n-1] = arr[n-1];
    for (int i=n-2; i>=0; i--)
        suffix[i] = suffix[i+1] + arr[i];

    // Storing all indexes with suffix
    // sum equal to total sum by 3.
    vector <int> v;
    for (int i=0; i<n; i++)
        if (suffix[i] == total_sum/3)
            v.push_back(i);

    // Traversing the prefix array and
    // doing binary search in above vector
    for (int i=0; i<n; i++)
    {
        // We found a prefix with total_sum/3
        if (prefix[i] == total_sum/3)
        {
            // Find first index in v[] which
            // is greater than i+1.
            int res = binarysearch(v, i+1);

            if (res != -1)
                count += v.size() - res;
        }
    }

    return count;
}

// driver function
int main()
{
    int arr[] = {1 , 2 , 3 , 0 , 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << CountContiguousParts(arr, n);
    return 0;
}

Output:

2

Time Complexity is O(n log n)

 

Efficient Approach [ O(n) solution ] :

  1. If sum of all the elements of the array is not divisible by 3, return 0.
  2. It is obvious that the sum of each part of each contiguous part will be equal to the sum of all elements divided by 3.
  3. Let’s create an array cnt[ ], where cnt[ i ] equals 1, if the sum of elements from i-th to n-th equals Array_Sum/3 else 0. Now, calculate the cumulative sum of the cnt array from the last index.
  4. Thus, we receive very simple solution: for each prefix of initial array 1…i with the sum that equals Array_Sum/3 we need to add to the answer sums[ i+2 ].

Below is the C++ code for the above approach.

// C++ program to count the ways
// to break the array in 3 equal parts
// having equal sum.
#include <bits/stdc++.h>
using namespace std;

// Function to count the no of ways
int countways(int a[], int n)
{
    int cnt[n] = {0}, s = 0;
    
    // Calculating the sum of the array
    // and storing it in variable s
    for (int i = 0 ; i < n ; i++)
    {
        s += a[i];
    }

    // Checking s is divisible by 3 or not
    if (s % 3 != 0)
        return 0;
    
    // Calculating the sum of each part
    s /= 3;
    
    int ss = 0;
    
    // If the sum of elements from i-th to n-th
    // equals to sum of part putting 1 in cnt 
    // array otherwise 0. 
    for (int i = n-1; i >= 0 ; i--)
    {
        ss += a[i];
        if (ss == s)
            cnt[i] = 1;
    }
    
    // Calculating the cumulative sum
    // of the array cnt from the last index.
    for (int i = n-2 ; i >= 0 ; i--)
        cnt[i] += cnt[i + 1];
    
    int ans = 0;
    ss = 0;
    
    // Calculating answer using original
    // and cnt array.
    for (int i = 0 ; i+2 < n ; i++)
    {
        ss += a[i];
        if (ss == s)
            ans += cnt[i + 2];
    }
    return ans;
}

// Driver function
int main()
{
    int n = 5;
    int a[] = {1, 2, 3, 0, 3};
    cout << countways(a, n) << endl;
    return 0;
}

Output:

2

Time Complexity: O(n).
This approach is contributed by Abhishek Sharma.

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:



3.2 Average Difficulty : 3.2/5.0
Based on 7 vote(s)