# Subsequences of size three in an array whose sum is divisible by m

Last Updated : 28 Mar, 2023

Given an array A[] (1<=A<=10) of size N (1<=N<=10), find the number of subsequences of length 3 whose sum is divisible by M (1<=M<=10).

Examples:

```Input  : A[] = {1, 2, 4, 3}
M = 3
Output : 2
Explanation : We can choose two such
subsequence of length 3 such that its
sum is divisible by 3 -> {1, 2, 3} ans
{2, 4, 3} ```

Brute Force Approach: We use 3 nested loops to traverse all the subsequences of length 3 to count all such subsequence which satisfies the given condition.
Here the brute force solution will work in O(N

Implementation:

## C++

 `// Brute Force Implementation (Time Complexity :``// O(N^3)) C++ program to find count of subsequences of size``// three divisible by M.``#include ` `using` `namespace` `std;` `int` `coutSubSeq(``int` `A[], ``int` `N, ``int` `M)``{``    ``int` `sum = 0;``    ``int` `ans = 0;` `    ``// Three nested loop to find all the sub ``    ``// sequences of length three in the given ``    ``// array A[].``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = i + 1; j < N; j++) {``            ``for` `(``int` `k = j + 1; k < N; k++) {``                ``sum = A[i] + A[j] + A[k];` `                ``// checking if the sum of the chosen``                ``// three number is divisible by m.``                ``if` `(sum % M == 0)``                    ``ans++;``            ``}``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `M = 3;``    ``int` `A[] = { 1, 2, 4, 3 };``    ``int` `N = ``sizeof``(A)/``sizeof``(A[0]);``    ``cout << coutSubSeq(A, N, M);``    ``return` `0;``}`

## Java

 `// Java program to find count of``// subsequences of size three``// divisible by M.``import` `java.io.*;` `class` `GFG {``    ` `    ``static` `int` `coutSubSeq(``int` `A[], ``int` `N,``                                   ``int` `M)``    ``{``        ``int` `sum = ``0``;``        ``int` `ans = ``0``;``     ` `        ``// Three nested loop to find all the``        ``// sub sequences of length three in``        ``// the given array A[].``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``for` `(``int` `j = i + ``1``; j < N; j++) {``                ``for` `(``int` `k = j + ``1``; k < N; k++) {``                    ``sum = A[i] + A[j] + A[k];``     ` `                    ``// checking if the sum of the``                    ``// chosen three number is ``                    ``// divisible by m.``                    ``if` `(sum % M == ``0``)``                        ``ans++;``                ``}``            ``}``        ``}``        ``return` `ans;``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `M = ``3``;``        ``int` `A[] = { ``1``, ``2``, ``4``, ``3` `};``        ``int` `N = A.length;``        ``System.out.println(coutSubSeq(A, N, M));``    ``}``}` `/*This code is contributed by Nikita Tiwari*/`

## Python

 `# Python program to find count of``# subsequences of size three ``# divisible by M.` `def` `coutSubSeq(A, N,M) :``    ``sum` `=` `0``    ``ans ``=` `0`` ` `    ``# Three nested loop to find all the``    ``# sub sequences of length three in ``    ``# the given array A[].``    ``for` `i ``in` `range``(``0``, N) :``        ``for` `j ``in` `range``(i ``+` `1``, N) :``            ``for` `k ``in` `range``(j ``+` `1``, N) :``                ``sum` `=` `A[i] ``+` `A[j] ``+` `A[k]`` ` `                ``# checking if the sum of ``                ``# the chosen three number``                ``# is divisible by m.``                ``if` `(``sum` `%` `M ``=``=` `0``) :``                    ``ans ``=` `ans ``+` `1``    ``return` `ans``    ` `# Driver code``M ``=` `3``A ``=` `[ ``1``, ``2``, ``4``, ``3` `]``N ``=` `len``(A)``print` `coutSubSeq(A, N, M)` `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find count of subsequences ``// of size three divisible by M.``using` `System;` `class` `GFG {``    ` `    ``static` `int` `coutSubSeq(``int``[] A, ``int` `N, ``int` `M)``    ``{``        ``int` `sum = 0;``        ``int` `ans = 0;``    ` `        ``// Three nested loop to find all the``        ``// sub sequences of length three in``        ``// the given array A[].``        ``for` `(``int` `i = 0; i < N; i++) {``            ``for` `(``int` `j = i + 1; j < N; j++) {``                ``for` `(``int` `k = j + 1; k < N; k++)``                ``{``                    ``sum = A[i] + A[j] + A[k];``    ` `                    ``// checking if the sum of``                    ``// the chosen three number``                    ``// is divisible by m.``                    ``if` `(sum % M == 0)``                        ``ans++;``                ``}``            ``}``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `M = 3;``        ``int` `[]A = { 1, 2, 4, 3 };``        ``int` `N = A.Length;``        ``Console.WriteLine(coutSubSeq(A, N, M));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output
`2`

Second Approach (Efficient for small M):

The above implementation takes a lot of time. So we need a better approach to solve this. Here we will use index mapping (hashing technique) to solve the problem. As we need to check if the sum of the chosen three numbers is divisible by M, we store each value under modulo M. We use a frequency array to store the number of occurrences of each element using index mapping techniques.

Three cases may occur:

• Firstly, if thrice of a number is divisible by M then we will add to answer where N is frequency of that number.
• Secondly, if twice of some number A added with sum number B is divisible by M then we will add * Freq[B] to the answer.
• Thirdly, if all the numbers A, B, C that sum up to S which is divisible by M then we will add Freq[A] * Freq[B] * Freq[C] to the answer.

With this approach, we can solve the problem in O(M) which is executable in the given time.

Implementation:

## C++

 `// O(M^2) time complexity C++ program to find ``// count of subsequences of size three``// divisible by M.``#include ` `using` `namespace` `std;` `int` `countSubSeq(``int` `A[], ``int` `N, ``int` `M)``{``    ``int` `ans = 0;` `    ``// Storing frequencies of all remainders``    ``// when divided by M.``    ``int` `h[M] = { 0 };``    ``for` `(``int` `i = 0; i < N; i++) {``        ``A[i] = A[i] % M;``        ``h[A[i]]++;``    ``}` `    ``for` `(``int` `i = 0; i < M; i++) {``        ``for` `(``int` `j = i; j < M; j++) {` `            ``// including i and j in the sum rem``            ``// calculate the remainder required ``            ``// to make the sum divisible by M``            ``int` `rem = (M - (i + j) % M) % M;` `            ``// if the required number is less than``            ``// j, we skip as we have already calculated ``            ``// for that value before. As j here starts ``            ``// with i and rem is less than j.``            ``if` `(rem < j)``                ``continue``;` `            ``// if satisfies the first case.``            ``if` `(i == j && rem == j)``                ``ans += h[i] * (h[i] - 1) * (h[i] - 2) / 6;` `            ``// if satisfies the second case``            ``else` `if` `(i == j)``                ``ans += h[i] * (h[i] - 1) * h[rem] / 2;` `            ``else` `if` `(i == rem)``                ``ans += h[i] * (h[i] - 1) * h[j] / 2;``            ``else` `if` `(rem == j)``                ``ans += h[j] * (h[j] - 1) * h[i] / 2;` `            ``// if satisfies the third case``            ``else``                ``ans = ans + h[i] * h[j] * h[rem];``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{   ``    ``int` `M = 3;``    ``int` `A[] = { 1, 2, 4, 3 };``    ``int` `N = ``sizeof``(A)/``sizeof``(A[0]);``    ``cout << countSubSeq(A, N, M);``    ``return` `0;``}`

## Java

 `// Java program to find count of ``// subsequences of size three``// divisible by M.``import` `java.io.*;``import` `java.util.Arrays;` `class` `GFG {` `    ``static` `int` `countSubSeq(``int` `A[], ``int` `N, ``int` `M)``    ``{``        ``int` `ans = ``0``;``     ` `        ``// Storing frequencies of all ``        ``// remainders when divided by M.``        ``int` `h[] = ``new` `int``[M];``        ``Arrays.fill(h,``0``);``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``A[i] = A[i] % M;``            ``h[A[i]]++;``        ``}``     ` `        ``for` `(``int` `i = ``0``; i < M; i++) {``            ``for` `(``int` `j = i; j < M; j++) {``     ` `                ``// including i and j in the sum``                ``// rem calculate the remainder``                ``// required to make the sum``                ``// divisible by M``                ``int` `rem = (M - (i + j) % M) % M;``     ` `                ``// if the required number is ``                ``// less than j, we skip as we ``                ``// have already calculated for``                ``// that value before. As j here``                ``// starts with i and rem is ``                ``// less than j.``                ``if` `(rem < j)``                    ``continue``;``     ` `                ``// if satisfies the first case.``                ``if` `(i == j && rem == j)``                    ``ans += h[i] * (h[i] - ``1``) *``                           ``(h[i] - ``2``) / ``6``;``     ` `                ``// if satisfies the second case``                ``else` `if` `(i == j)``                    ``ans += h[i] * (h[i] - ``1``) *``                           ``h[rem] / ``2``;``     ` `                ``else` `if` `(i == rem)``                    ``ans += h[i] * (h[i] - ``1``) *``                           ``h[j] / ``2``;``                ``else` `if` `(rem == j)``                    ``ans += h[j] * (h[j] - ``1``) *``                           ``h[i] / ``2``;``     ` `                ``// if satisfies the third case``                ``else``                    ``ans = ans + h[i] * h[j] *``                          ``h[rem];``            ``}``        ``}``        ``return` `ans;``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `M = ``3``;``        ``int` `A[] = { ``1``, ``2``, ``4``, ``3` `};``        ``int` `N = A.length;``        ``System.out.println(countSubSeq(A, N, M));``    ``}``}` `/*This code is contributed by Nikita Tiwari*/`

## Python

 `# Python program to find count of``# subsequences of size three``# divisible by M.` `def` `countSubSeq(A, N, M) :``    ``ans ``=` `0``    ` `    ``# Storing frequencies of all``    ``# remainders when divided by M.``    ``h ``=` `[``0``] ``*` `M``    ``for` `i ``in` `range``(``0``,N) :``        ``A[i] ``=` `A[i] ``%` `M``        ``h[A[i]] ``=` `h[A[i]] ``+` `1``    ` `    ``for` `i ``in` `range``(``0``,M) :``        ``for` `j ``in` `range``(i,M) :``            ` `            ``# including i and j in the sum``            ``# rem calculate the remainder``            ``# required to make the sum``            ``# divisible by M``            ``rem ``=` `(M ``-` `(i ``+` `j) ``%` `M) ``%` `M``            ` `            ``# if the required number is ``            ``# less than j, we skip as we ``            ``# have already calculated for``            ``# that value before. As j here``            ``# starts with i and rem is ``            ``# less than j.``            ``if` `(rem < j) :``                ``continue``                ` `            ``# if satisfies the first case.``            ``if` `(i ``=``=` `j ``and` `rem ``=``=` `j) :``                ``ans ``=` `ans ``+` `h[i] ``*` `(h[i] ``-` `1``) ``*` `(h[i] ``-` `2``) ``/` `6``                ` `                ` `            ``# if satisfies the second case``            ``elif` `(i ``=``=` `j) :``                ``ans ``=` `ans ``+``( h[i] ``*` `(h[i] ``-` `1``) ``*` `h[rem] ``/` `2``)``     ` `            ``elif` `(i ``=``=` `rem) :``                ``ans ``=` `ans ``+` `h[i] ``*` `(h[i] ``-` `1``) ``*` `h[j] ``/` `2``            ``elif` `(rem ``=``=` `j) :``                ``ans ``=` `ans ``+` `h[j] ``*` `(h[j] ``-` `1``) ``*` `h[i] ``/` `2``     ` `            ``# if satisfies the third case``            ``else` `:``                ``ans ``=` `ans ``+` `h[i] ``*` `h[j] ``*` `h[rem]``            ` `        ``return` `ans``        ` `# Driver code``M ``=` `3``;``A ``=` `[ ``1``, ``2``, ``4``, ``3` `]``N ``=` `len``(A)``print``(countSubSeq(A, N, M))` `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find count of subsequences of``// size three divisible by M.``using` `System;` `class` `GFG {` `    ``static` `int` `countSubSeq(``int` `[]A, ``int` `N, ``int` `M)``    ``{``        ``int` `ans = 0;``    ` `        ``// Storing frequencies of all ``        ``// remainders when divided by M.``        ``int` `[]h = ``new` `int``[M];``    ` `        ``for` `(``int` `i = 0; i < N; i++) {``            ``A[i] = A[i] % M;``            ``h[A[i]]++;``        ``}``    ` `        ``for` `(``int` `i = 0; i < M; i++) {``            ``for` `(``int` `j = i; j < M; j++) {``    ` `                ``// including i and j in the sum``                ``// rem calculate the remainder``                ``// required to make the sum``                ``// divisible by M``                ``int` `rem = (M - (i + j) % M) % M;``    ` `                ``// if the required number is ``                ``// less than j, we skip as we ``                ``// have already calculated for``                ``// that value before. As j here``                ``// starts with i and rem is ``                ``// less than j.``                ``if` `(rem < j)``                    ``continue``;``    ` `                ``// if satisfies the first case.``                ``if` `(i == j && rem == j)``                    ``ans += h[i] * (h[i] - 1) *``                                ``(h[i] - 2) / 6;``    ` `                ``// if satisfies the second case``                ``else` `if` `(i == j)``                    ``ans += h[i] * (h[i] - 1) *``                                    ``h[rem] / 2;``    ` `                ``else` `if` `(i == rem)``                    ``ans += h[i] * (h[i] - 1) *``                                     ``h[j] / 2;``                ``else` `if` `(rem == j)``                    ``ans += h[j] * (h[j] - 1) *``                                     ``h[i] / 2;``    ` `                ``// if satisfies the third case``                ``else``                    ``ans = ans + h[i] * h[j] *``                                      ``h[rem];``            ``}``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `M = 3;``        ``int` `[]A = { 1, 2, 4, 3 };``        ``int` `N = A.Length;``        ``Console.WriteLine(countSubSeq(A, N, M));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output
`2`

complexity analysis:

The time complexity of the given program is O(M^2) as it has two nested loops iterating over the range [0, M)
The space complexity is O(M) as an array of size M is created to store the frequencies of remainders when divided by M.

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