# Count of subsequences with sum two less than the array sum

• Last Updated : 12 Dec, 2021

Given an array vec[] of size N of non-negative integers. The task is to count the number of subsequences with the sum equal to S – 2 where S is the sum of all the elements of the array.

Examples:

Input: vec[] = {2, 0, 1, 2, 1}, N=5
Output: 6
Explanation: {2, 0, 1, 1}, {2, 1, 1}, {2, 0, 2}, {2, 2}, {0, 1, 2, 1}, {1, 2, 1}

Input: vec[] = {2, 0, 2, 3, 1}, N=5
Output: 4
Explanation: {2, 0, 3, 1}, {2, 3, 1}, {0, 2, 3, 1}, {2, 3, 1}

Naive Approach: The idea is to generate all subsequences and check the sum of each and every individual subsequence equals S-2 or not.

Below is the implementation of the above approach.

## C++

 // C++ program for the above approach#include using namespace std; // Function to count the total number// of subsequences with sum S-2void countTotal(vector& vec){     // Calculating vector sum using    // accumulate function    int sum = accumulate(vec.begin(),                         vec.end(), 0LL);     int N = (int)vec.size();     // Answer variable stores count    // of subsequences with desired sum    int answer = 0;     // Bitmasking technique to generate    // all possible subsequences    for (int mask = 0; mask < (1 << N); mask++) {         // Variable curr counts the        // sum of the current subsequence        int curr = 0;        for (int i = 0; i < N; i++) {             if ((mask & (1 << i)) != 0) {                 // Include ith element                // of the vector                curr += vec[i];            }        }         if (curr == sum - 2)            answer++;    }     // Print the answer    cout << answer;} // Driver Codeint main(){     // Initializing a vector    vector vec = { 2, 0, 1, 2, 1 };     countTotal(vec);     return 0;}

## Java

 // Java program for the above approachpublic class GFG {         static int accumulate(int [] vec){        int sum1 = 0;                 for (int i = 0; i < vec.length; i++)            sum1 += vec[i];    return sum1;    }         // Function to count the total number    // of subsequences with sum S-2    static void countTotal(int []vec)    {             // Calculating vector sum using        // accumulate function        int sum = accumulate(vec);             int N = vec.length;             // Answer variable stores count        // of subsequences with desired sum        int answer = 0;             // Bitmasking technique to generate        // all possible subsequences        for (int mask = 0; mask < (1 << N); mask++) {                 // Variable curr counts the            // sum of the current subsequence            int curr = 0;            for (int i = 0; i < N; i++) {                     if ((mask & (1 << i)) != 0) {                         // Include ith element                    // of the vector                    curr += vec[i];                }            }                 if (curr == sum - 2)                answer++;        }             // Print the answer        System.out.print(answer);    }         // Driver Code    public static void main (String[] args)    {             // Initializing a vector        int []vec = { 2, 0, 1, 2, 1 };             countTotal(vec);    } } // This code is contributed by AnkThon

## Python3

 # python3 program for the above approach # Function to count the total number# of subsequences with sum S-2def countTotal(vec) :     # Calculating vector sum using    # accumulate function    Sum = sum(vec)     N = len(vec);     # Answer variable stores count    # of subsequences with desired sum    answer = 0;     # Bitmasking technique to generate    # all possible subsequences    for mask in range((1 << N)) :                 # Variable curr counts the        # sum of the current subsequence        curr = 0;        for i in range(N) :             if ((mask & (1 << i)) != 0) :                 # Include ith element                # of the vector                curr += vec[i];                         if (curr == Sum - 2) :            answer += 1;     # Print the answer    print(answer); # Driver Codeif __name__ == "__main__" :      # Initializing a vector    vec = [ 2, 0, 1, 2, 1 ];     countTotal(vec);     # This code is contributed by AnkThon

## C#

 // C# program for the above approachusing System; public class GFG {     static int accumulate(int[] vec)    {        int sum1 = 0;         for (int i = 0; i < vec.Length; i++)            sum1 += vec[i];        return sum1;    }     // Function to count the total number    // of subsequences with sum S-2    static void countTotal(int[] vec)    {         // Calculating vector sum using        // accumulate function        int sum = accumulate(vec);         int N = vec.Length;         // Answer variable stores count        // of subsequences with desired sum        int answer = 0;         // Bitmasking technique to generate        // all possible subsequences        for (int mask = 0; mask < (1 << N); mask++) {             // Variable curr counts the            // sum of the current subsequence            int curr = 0;            for (int i = 0; i < N; i++) {                 if ((mask & (1 << i)) != 0) {                     // Include ith element                    // of the vector                    curr += vec[i];                }            }             if (curr == sum - 2)                answer++;        }         // Print the answer        Console.WriteLine(answer);    }     // Driver Code    public static void Main(string[] args)    {         // Initializing a vector        int[] vec = { 2, 0, 1, 2, 1 };         countTotal(vec);    }} // This code is contributed by ukasp.

## Javascript



Output

6

Time Complexity: O(2N
Auxiliary Space: O(1)

Efficient Approach: The idea is to use the Combinatorics that apart from 0’s, 1’s, and 2’s, all the other elements in our array will be part of the desired subsequences. Let’s call them extra elements. Then, count the occurrences of 0’s, 1’s, and 2’s in the array. Let’s say the count of 0’s is x, count of 1’s be y, count of 2’s be z

• Let’s count the number of desired subsequences if all 2’s and extra elements are in the subsequence. Now there can be exactly y – 2 elements out of y. Note that there is no restriction for taking 0’s as it contributes nothing to our subsequence sum.
• Hence, the total count of such subsequences = count1 = 2x Ã— yCy – 2  = 2x Ã— yC2 ( Since, nC0 + nC1 + . . . + nCn  = 2n).
• Let’s count the number of subsequences if all the 1’s are in our subsequence. Now there can be exactly z – 1 elements out of z.
• Hence, the total count of such subsequences = count2 = 2 Ã— zCz – 1 = 2x  Ã—   zC1
• Total count of subsequences whose sum is equal to S – 2, count = count1 + count2 = 2x Ã— ( yC2 + zC1 )

Follow the steps below to solve the problem:

• Initialize the variable sum as the sum of the array.
• Initialize the variables countOfZero, countOfOne and countOfTwo to store the count of 0, 1 and 2.
• Traverse the array vec[] using the iterator x and perform the following tasks:
• Count the occurrences of 0’s, 1’s, and 2’s.
• Initialize the variables value1 as 2countOfZero.
• Initialize the variable value2 as (countOfOne * (countOfOne – 1)) / 2.
• Initialize the variable value3 as countOfTwo.
• Set the value of answer as value1 * ( value2 + value).
• After performing the above steps, print the value of answer as the answer.

Below is the implementation of the above approach.

## C++

 // C++ program for the above approach#include using namespace std; // Function to count the total number// of subsequences with sum S-2void countTotal(vector& vec){     // Calculating vector sum using    // accumulate function    int sum = accumulate(vec.begin(),                         vec.end(), 0LL);     int N = (int)vec.size();     // Answer variable stores count    // of subsequences with desired sum    int answer = 0;     int countOfZero = 0,        countOfOne = 0,        countOfTwo = 0;    for (auto x : vec) {        if (x == 0)            countOfZero++;        else if (x == 1)            countOfOne++;        else if (x == 2)            countOfTwo++;    }     // Select any number of    // zeroes from 0 to count[0]    // which is equivalent    // to 2 ^ countOfZero    int value1 = pow(2, countOfZero);     // Considering all 2's    // and extra elements    int value2        = (countOfOne           * (countOfOne - 1))          / 2;     // Considering all 1's    // and extra elements    int value3 = countOfTwo;     // Calculating final answer    answer = value1 * (value2 + value3);     // Print the answer    cout << answer;} // Driver Codeint main(){     // Initializing a vector    vector vec = { 2, 0, 1, 2, 1 };     countTotal(vec);     return 0;}

## Java

 /*package whatever //do not write package name here */import java.io.*; class GFG {    // Function to count the total number    // of subsequences with sum S-2    static void countTotal(int[] arr)    {         int N = arr.length;         // Answer variable stores count        // of subsequences with desired sum        int answer = 0;         int countOfZero = 0, countOfOne = 0, countOfTwo = 0;        for (int i = 0; i < N; i++) {             if (arr[i] == 0)                countOfZero++;            else if (arr[i] == 1)                countOfOne++;            else if (arr[i] == 2)                countOfTwo++;        }         // Select any number of        // zeroes from 0 to count[0]        // which is equivalent        // to 2 ^ countOfZero        int value1 = (1 << countOfZero);         // Considering all 2's        // and extra elements        int value2 = (countOfOne * (countOfOne - 1)) / 2;         // Considering all 1's        // and extra elements        int value3 = countOfTwo;         // Calculating final answer        answer = value1 * (value2 + value3);         // Print the answer        System.out.println(answer);    }     // Driver Code    public static void main(String[] args)    {        // Initializing an array        int[] arr = { 2, 0, 1, 2, 1 };         countTotal(arr);    }} // This code is contributed by maddler.

## Python3

 # Python3 program for the above approach # Function to count the total number# of subsequences with sum S-2def countTotal(vec) :     # Calculating vector sum using    # accumulate function    sum1 = sum(vec);     N = len(vec);     # Answer variable stores count    # of subsequences with desired sum    answer = 0;    countOfZero = 0; countOfOne = 0; countOfTwo = 0;    for x in vec :                 if (x == 0) :            countOfZero += 1;                     elif (x == 1) :            countOfOne += 1;                     elif (x == 2) :            countOfTwo += 1;     # Select any number of    # zeroes from 0 to count[0]    # which is equivalent    # to 2 ^ countOfZero    value1 = 2 ** countOfZero;     # Considering all 2's    # and extra elements    value2 = (countOfOne * (countOfOne - 1)) // 2;     # Considering all 1's    # and extra elements    value3 = countOfTwo;     # Calculating final answer    answer = value1 * (value2 + value3);     # Print the answer    print(answer); # Driver Codeif __name__ == "__main__" :     # Initializing a vector    vec = [ 2, 0, 1, 2, 1 ];    countTotal(vec);         # This code is contributed by AnkThon

## C#

 // Above approach implemented in C#using System; public class GFG {         // Function to count the total number    // of subsequences with sum S-2    static void countTotal(int[] arr)    {         int N = arr.Length;         // Answer variable stores count        // of subsequences with desired sum        int answer = 0;         int countOfZero = 0, countOfOne = 0, countOfTwo = 0;        for (int i = 0; i < N; i++) {             if (arr[i] == 0)                countOfZero++;                             else if (arr[i] == 1)                countOfOne++;                             else if (arr[i] == 2)                countOfTwo++;        }         // Select any number of        // zeroes from 0 to count[0]        // which is equivalent        // to 2 ^ countOfZero        int value1 = (1 << countOfZero);         // Considering all 2's        // and extra elements        int value2 = (countOfOne * (countOfOne - 1)) / 2;         // Considering all 1's        // and extra elements        int value3 = countOfTwo;         // Calculating final answer        answer = value1 * (value2 + value3);         // Print the answer        Console.WriteLine(answer);    }     // Driver Code    public static void Main(string[] args)    {               // Initializing an array        int[] arr = { 2, 0, 1, 2, 1 };         countTotal(arr);    }} // This code is contributed by AnkThon

## Javascript



Output

6

Time Complexity: O(N)
Auxiliary Space: O(1)

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