Check if an array can be split into 3 subsequences of equal sum or not

Check if an array can be split into 3 subsequences of equal sum or not

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Given an array arr[] having N integers. The task is to determine if the array can be partitioned into 3 subsequences of equal sum or not. If yes then print “Yes”. Otherwise, print “No”.

Examples:

Input:_ arr[] = {1, 1, 1}_

Output:_ Yes_

Explanation:

_Here array can be partition into 3 equal sum. {1}  _

Input:_ arr[] = {40}_

Output:_ No_

Explanation:

Here array cannot be partition into 3 equal sum.

Recursive Approach: This problem can be solved using recursion. Below are the steps:

  1. Initialize three variable sum1sum2, and sum3 to value 0.
  2. Then every element of array arr[] is added to either of these 3 variables, which give all the possible combinations.
  3. In case, any subsequences having 3 equal sums then the array can be partition.
  4. If the array can be partition then print “Yes” else print “No”.
  • C++

// C++ program for the above approach

#include <bits/stdc++.h>

**using** **namespace** std;

// Utility function to check array can

// be partition to 3 subsequences of

// equal sum or not

**int** checkEqualSumUtil(``**int** arr[], **int** N,

**int** sm1, **int** sm2,

**int** sm3, **int** j)

{

// Base Case

**if** (j == N) {

**if** (sm1 == sm2 && sm2 == sm3)

**return** 1;

**else**

**return** 0;

}

**else** {

// When element at index

// j is added to sm1

**int** l = checkEqualSumUtil(

arr, N,

sm1 + arr[j], sm2,

sm3, j + 1);

// When element at index

// j is added to sm2

**int** m = checkEqualSumUtil(

arr, N, sm1,

sm2 + arr[j],

sm3, j + 1);

// When element at index

// j is added to sm3

**int** r = checkEqualSumUtil(

arr, N, sm1, sm2,

sm3 + arr[j], j + 1);

// Return maximum value among

// all above 3 recursive call

**return** max(max(l, m), r);

}

}

// Function to check array can be

// partition to 3 subsequences of

// equal sum or not

**void** checkEqualSum(``**int** arr[], **int** N)

{

// Initialise 3 sums to 0

**int** sum1, sum2, sum3;

sum1 = sum2 = sum3 = 0;

// Function Call

**if** (checkEqualSumUtil(

arr, N, sum1,

sum2, sum3, 0)

== 1) {

cout << "Yes"``;

}

**else** {

cout << "No"``;

}

}

// Driver Code

**int** main()

{

// Given array arr[]

**int** arr[]

= { 17, 34, 59, 23, 17, 67,

57, 2, 18, 59, 1 };

**int** N = **sizeof**``(arr) / **sizeof**``(arr[0]);

// Function Call

checkEqualSum(arr, N);

**return** 0;

}

Output:

Yes

Time Complexity:_ O(3N)_

Auxiliary Space:_ O(1)_

Dynamic Programming** Approach:** This problem can be solved using dynamic programming, the idea is to store all the overlapping subproblems value in a map and use the value of overlapping substructure to reduce the number of the recursive calls. Below are the steps:

  1. Let sum1sum2, and sum3 be the three equal sum to be partitioned.
  2. Create a map dp having the key:

(to_string(sum1) + “_” + _**_to_string_(sum2) + “_” + _to_string**(sum3))

  1. with value is 1 if 3 equal subsets are found else value is 0.
  2. Traverse the given array and do the following:
  • Base Case: While traversing the array if we reach the end of the array then check if the value of sum1sum2, and sum3 are equal then return 1 that will ensure that we can break the given array into subsequence of equal sum value. Otherwise return 0.
  • Recursive Call: For each element in the array include each element in sum1sum2, and sum3 one by one and return the maximum of the these recursive call.

a = recursive_function(arr, N, sum1 + arr[j], sum2, sum3, j + 1)

b = recursive_function(arr, N, sum1, sum2 + arr[j], sum3, j + 1)

c = recursive_function(arr, N, sum1, sum2, sum3 + arr[j], j + 1)

  • Return Statement: In the above recursive call the maximum of the three values will give the result for the current recursive call. Update the current state in the dp table as:

string s = to_string(sum1) + ‘_’ + to_string(sum1) + to_string(sum1)

return dp[s] = max(a, max(b, c)

  1. If there can be partition possible then print “Yes” else print “No”.

Below is the implementation of the above approach:

  • C++

// C++ program for the above approach

#include <bits/stdc++.h>

**using** **namespace** std;

map<string, **int**``> dp;

// Function to check array can be

// partition into sum of 3 equal

**int** checkEqualSumUtil(``**int** arr[], **int** N,

**int** sm1,

**int** sm2,

**int** sm3, **int** j)

{

string s = to_string(sm1)

+ "_" + to_string(sm2)

+ to_string(j);

// Base Case

**if** (j == N) {

**if** (sm1 == sm2 && sm2 == sm3)

**return** 1;

**else**

**return** 0;

}

// If value at particular index is not

// -1 then return value at that index

// which ensure no more further calls

**if** (dp.find(s) != dp.end())

**return** dp[s];

**else** {

// When element at index

// j is added to sm1

**int** l = checkEqualSumUtil(

arr, N, sm1 + arr[j],

sm2, sm3, j + 1);

// When element at index

// j is added to sm2

**int** m = checkEqualSumUtil(

arr, N, sm1, sm2 + arr[j],

sm3, j + 1);

// When element at index

// j is added to sm3

**int** r = checkEqualSumUtil(

arr, N, sm1, sm2,

sm3 + arr[j], j + 1);

// Update the current state and

// return that value

**return** dp[s] = max(max(l, m), r);

}

}

// Function to check array can be

// partition to 3 subsequences of

// equal sum or not

**void** checkEqualSum(``**int** arr[], **int** N)

{

// Initialise 3 sums to 0

**int** sum1, sum2, sum3;

sum1 = sum2 = sum3 = 0;

// Function Call

**if** (checkEqualSumUtil(

arr, N, sum1,

sum2, sum3, 0)

== 1) {

cout << "Yes"``;

}

**else** {

cout << "No"``;

}

}

// Driver Code

**int** main()

{

// Given array arr[]

**int** arr[]

= { 17, 34, 59, 23, 17, 67,

57, 2, 18, 59, 1 };

**int** N = **sizeof**``(arr) / **sizeof**``(arr[0]);

// Function Call

checkEqualSum(arr, N);

**return** 0;

}

Output:

Yes

Time Complexity:_ O(N*K2) _

Auxiliary Space:_ O(N*K2) where K is the sum of the array._

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