Count of subarrays which start and end with the same element

Given an array A of size N where the array elements contain values from 1 to N with duplicates, the task is to find total number of subarrays which start and end with the same element.

Examples:

Input: A[] = {1, 2, 1, 5, 2}
Output: 7
Explanation:
Total 7 sub-array of the given array are {1}, {2}, {1}, {5}, {2}, {1, 2, 1} and {2, 1, 5, 2} are start and end with same element.

Input: A[] = {1, 5, 6, 1, 9, 5, 8, 10, 8, 9}
Output: 14
Explanation:
Total 14 sub-array {1}, {5}, {6}, {1}, {9}, {5}, {8}, {10}, {8}, {9}, {1, 5, 6, 1}, {5, 6, 1, 9, 5}, {9, 5, 8, 10, 8, 9} and {8, 10, 8} start and end with same element.

Naive approach: For each element in the array, if it is present at a different index as well, we will increase our result by 1. Also, all 1-size subarray are part of counted in the result. Therefore, add N to the result.



Below is the implementation of the above approach:

C++

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// C++ program to Count total sub-array
// which start and end with same element
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find total sub-array
// which start and end with same element
void cntArray(int A[], int N)
{
    // initialize result with 0
    int result = 0;
  
    for (int i = 0; i < N; i++) {
  
        // all size 1 sub-array
        // is part of our result
        result++;
  
        // element at current index
        int current_value = A[i];
  
        for (int j = i + 1; j < N; j++) {
  
            // Check if A[j] = A[i]
            // increase result by 1
            if (A[j] == current_value) {
                result++;
            }
        }
    }
  
    // print the result
    cout << result << endl;
}
  
// Driver code
int main()
{
    int A[] = { 1, 5, 6, 1, 9,
                5, 8, 10, 8, 9 };
    int N = sizeof(A) / sizeof(int);
  
    cntArray(A, N);
  
    return 0;
}

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Java

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// Java program to Count total sub-array
// which start and end with same element
  
public class Main {
  
    // function to find total sub-array
    // which start and end with same element
    public static void cntArray(int A[], int N)
    {
        // initialize result with 0
        int result = 0;
  
        for (int i = 0; i < N; i++) {
  
            // all size 1 sub-array
            // is part of our result
            result++;
  
            // element at current index
            int current_value = A[i];
  
            for (int j = i + 1; j < N; j++) {
  
                // Check if A[j] = A[i]
                // increase result by 1
                if (A[j] == current_value) {
                    result++;
                }
            }
        }
  
        // print the result
        System.out.println(result);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[] A = { 1, 5, 6, 1, 9,
                    5, 8, 10, 8, 9 };
        int N = A.length;
        cntArray(A, N);
    }
}

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Python3

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# Python3 program to count total sub-array 
# which start and end with same element 
  
# Function to find total sub-array 
# which start and end with same element 
def cntArray(A, N): 
  
    # Initialize result with 0 
    result = 0
  
    for i in range(0, N): 
  
        # All size 1 sub-array 
        # is part of our result 
        result = result + 1
  
        # Element at current index 
        current_value = A[i]
  
        for j in range(i + 1, N): 
  
            # Check if A[j] = A[i] 
            # increase result by 1 
            if (A[j] == current_value): 
                result = result + 1
  
    # Print the result 
    print(result)
    print("\n")
  
# Driver code 
A = [ 1, 5, 6, 1, 9, 5, 8, 10, 8, 9
N = len(A)
  
cntArray(A, N)
  
# This code is contributed by PratikBasu 

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C#

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// C# program to Count total sub-array
// which start and end with same element
using System;
class GFG{
  
// function to find total sub-array
// which start and end with same element
public static void cntArray(int []A, int N)
{
    // initialize result with 0
    int result = 0;
  
    for (int i = 0; i < N; i++) 
    {
  
        // all size 1 sub-array
        // is part of our result
        result++;
  
        // element at current index
        int current_value = A[i];
  
        for (int j = i + 1; j < N; j++) 
        {
  
            // Check if A[j] = A[i]
            // increase result by 1
            if (A[j] == current_value) 
            {
                result++;
            }
        }
    }
  
    // print the result
    Console.Write(result);
}
  
// Driver code
public static void Main()
{
    int[] A = { 1, 5, 6, 1, 9,
                5, 8, 10, 8, 9 };
    int N = A.Length;
    cntArray(A, N);
}
}
  
// This code is contributed by Code_Mech

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Output:

14

Time Complexity: O(N2), where N is the size of array

Space complexity: O(1)

Efficient approach: We can optimize the above method by observing that the answer just depends on frequencies of numbers in the original array.
For example in array {1, 5, 6, 1, 9, 5, 8, 10, 8, 9}, frequency of 1 is 2 and sub-array contributing to answer are {1}, {1} and {1, 5, 6, 1} respectively, i.e., a total of 3.
Therefore calculate the frequency of each element in the array. Then for each element, the increment the count by the result yielded by the following formula:

( (frequency of element)*(frequency of element + 1) ) / 2

Below is the implementation of the above approach:

C++

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// C++ program to Count total sub-array
// which start and end with same element
  
#include <bits/stdc++.h>
using namespace std;
  
// function to find total sub-array
// which start and end with same element
void cntArray(int A[], int N)
{
    // initialize result with 0
    int result = 0;
  
    // array to count frequency of 1 to N
    int frequency[N + 1] = { 0 };
  
    for (int i = 0; i < N; i++) {
        // update frequency of A[i]
        frequency[A[i]]++;
    }
  
    for (int i = 1; i <= N; i++) {
        int frequency_of_i = frequency[i];
  
        // update result with sub-array
        // contributed by number i
        result += +((frequency_of_i)
                    * (frequency_of_i + 1))
                  / 2;
    }
  
    // print the result
    cout << result << endl;
}
  
// Driver code
int main()
{
    int A[] = { 1, 5, 6, 1, 9, 5,
                8, 10, 8, 9 };
    int N = sizeof(A) / sizeof(int);
  
    cntArray(A, N);
  
    return 0;
}

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Java

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// Java program to Count total sub-array
// which start and end with same element
  
public class Main {
  
    // function to find total sub-array which
    // start and end with same element
    public static void cntArray(int A[], int N)
    {
        // initialize result with 0
        int result = 0;
  
        // array to count frequency of 1 to N
        int[] frequency = new int[N + 1];
  
        for (int i = 0; i < N; i++) {
            // update frequency of A[i]
            frequency[A[i]]++;
        }
  
        for (int i = 1; i <= N; i++) {
            int frequency_of_i = frequency[i];
  
            // update result with sub-array
            // contributed by number i
            result += ((frequency_of_i)
                       * (frequency_of_i + 1))
                      / 2;
        }
  
        // print the result
        System.out.println(result);
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        int[] A = { 1, 5, 6, 1, 9, 5,
                    8, 10, 8, 9 };
        int N = A.length;
        cntArray(A, N);
    }
}

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Python3

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# Python3 program to count total sub-array 
# which start and end with same element 
  
# Function to find total sub-array 
# which start and end with same element 
def cntArray(A, N): 
  
    # Initialize result with 0 
    result = 0
  
    # Array to count frequency of 1 to N 
    frequency = [0] * (N + 1
  
    for i in range(0, N):
          
        # Update frequency of A[i] 
        frequency[A[i]] = frequency[A[i]] + 1
  
    for i in range(1, N + 1): 
        frequency_of_i = frequency[i]
  
        # Update result with sub-array 
        # contributed by number i 
        result = result + ((frequency_of_i) *
                           (frequency_of_i + 1)) / 2
  
    # Print the result 
    print(int(result))
    print("\n")
  
# Driver code 
A = [ 1, 5, 6, 1, 9, 5, 8, 10, 8, 9 ]
N = len(A)
  
cntArray(A, N) 
  
# This code is contributed by PratikBasu 

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C#

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// C# program to Count total sub-array
// which start and end with same element
using System;
class GFG{
  
// function to find total sub-array which
// start and end with same element
public static void cntArray(int []A, int N)
{
    // initialize result with 0
    int result = 0;
  
    // array to count frequency of 1 to N
    int[] frequency = new int[N + 1];
  
    for (int i = 0; i < N; i++) 
    {
        // update frequency of A[i]
        frequency[A[i]]++;
    }
  
    for (int i = 1; i <= N; i++) 
    {
        int frequency_of_i = frequency[i];
  
        // update result with sub-array
        // contributed by number i
        result += ((frequency_of_i) * 
                   (frequency_of_i + 1)) / 2;
    }
  
    // print the result
    Console.Write(result);
}
  
// Driver code
public static void Main()
{
    int[] A = { 1, 5, 6, 1, 9, 5,
                8, 10, 8, 9 };
    int N = A.Length;
    cntArray(A, N);
}
}
  
// This code is contributed by Nidhi_Biet

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Output:

14

Time Complexity: O(N), where N is the size of array

Space complexity: O(N)

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