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Count of subarrays which start and end with the same element

  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021

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

Examples: 

Input: A[] = {1, 2, 1, 5, 2} 
Output:
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++




// 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;
}

Java




// 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);
    }
}

Python3




# 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

C#




// 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

Javascript




<script>
 
// Javascript 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
function cntArray(A, N)
{
    // initialize result with 0
    var result = 0;
 
    for (var i = 0; i < N; i++) {
 
        // all size 1 sub-array
        // is part of our result
        result++;
 
        // element at current index
        var current_value = A[i];
 
        for (var 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
    document.write( result );
}
 
// Driver code
var A = [1, 5, 6, 1, 9,
            5, 8, 10, 8, 9];
var N = A.length;
cntArray(A, N);
 
// This code is contributed by noob2000.
</script>
Output: 
14

 

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




// 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;
}

Java




// 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);
    }
}

Python3




# 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

C#




// 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

Javascript




<script>
 
// Javascript 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
    function cntArray(A, N)
    {
        // initialize result with 0
        let result = 0;
   
        // array to count frequency of 1 to N
        let frequency = Array.from({length: N+1}, (_, i) => 0);
   
        for (let i = 0; i < N; i++) {
            // update frequency of A[i]
            frequency[A[i]]++;
        }
   
        for (let i = 1; i <= N; i++) {
            let frequency_of_i = frequency[i];
   
            // update result with sub-array
            // contributed by number i
            result += ((frequency_of_i)
                       * (frequency_of_i + 1))
                      / 2;
        }
   
        // prlet the result
        document.write(result);
    }
 
 
// Driver Code
     
    let A = [ 1, 5, 6, 1, 9, 5,
                    8, 10, 8, 9 ];
        let N = A.length;
        cntArray(A, N);
       
</script>
Output: 
14

 

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




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