Write a PHP program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum.

Examples:

Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True
Explanation: There is a subset (4, 5) with sum 9.

Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
Explanation: There is no subset that adds up to 30.

PHP Program for Subset Sum Problem using Recursion:

For the recursive approach, there will be two cases.

• Consider the ‘last’ element to be a part of the subset. Now the new required sum = required sum – value of ‘last’ element.
• Don’t include the ‘last’ element in the subset. Then the new required sum = old required sum.

In both cases, the number of available elements decreases by 1.

Step-by-step approach:

• Build a recursive function and pass the index to be considered (here gradually moving from the last end) and the remaining sum amount.
• For each index check the base cases and utilize the above recursive call.
• If the answer is true for any recursion call, then there exists such a subset. Otherwise, no such subset exists.

Below is the implementation of the above approach.

<?php
// A recursive solution for subset sum problem

// Returns true if there is a subset of set
// with sun equal to given sum
function isSubsetSum($set, $n, $sum)
{
    // Base Cases
    if ($sum == 0)
        return true;
    if ($n == 0)
        return false;
    
    // If last element is greater
    // than sum, then ignore it
    if ($set[$n - 1] > $sum)
        return isSubsetSum($set, $n - 1, $sum);
    
    // Else, check if sum can be 
    // obtained by any of the following
    // (a) including the last element
    // (b) excluding the last element
    return isSubsetSum($set, $n - 1, $sum) || 
        isSubsetSum($set, $n - 1, 
                    $sum - $set[$n - 1]);
}

// Driver Code
$set = array(3, 34, 4, 12, 5, 2);
$sum = 9;
$n = 6;

if (isSubsetSum($set, $n, $sum) == true)
    echo"Found a subset with given sum";
else
    echo "No subset with given sum";
    
// This code is contributed by Anuj_67 
?>

Found a subset with given sum

Time Complexity: O(2ⁿ)
Auxiliary space: O(n)

PHP Program for Subset Sum Problem using Memoization:

As seen in the previous recursion method, each state of the solution can be uniquely identified using two variables – the index and the remaining sum. So create a 2D array to store the value of each state to avoid recalculation of the same state.

Below is the implementation of the above approach:

<?php
// Check if possible subset with
// given sum is possible or not
function subsetSum($a, $n, $sum) {
    // Storing the value -1 to the matrix
    $tab = array();
    for ($i = 1; $i <= $n; $i++) {
        for ($j = 1; $j <= $sum; $j++) {
            $tab[$i][$j] = -1;
        }
    }

    // If the sum is zero, it means
    // we got our expected sum
    if ($sum == 0)
        return 1;

    if ($n <= 0)
        return 0;

    // If the value is not -1, it means it
    // already called the function
    // with the same value.
    // It will save us from repetition.
    if ($tab[$n - 1][$sum] != -1)
        return $tab[$n - 1][$sum];

    // If the value of a[n-1] is
    // greater than the sum.
    // We call for the next value.
    if ($a[$n - 1] > $sum)
        return $tab[$n - 1][$sum] = subsetSum($a, $n - 1, $sum);
    else {
        // Here we do two calls because we
        // don't know which value fulfills our criteria.
        // That's why we're doing two calls.
        if (subsetSum($a, $n - 1, $sum) != 0 || subsetSum($a, $n - 1, $sum - $a[$n - 1]) != 0) {
            return $tab[$n - 1][$sum] = 1;
        } else
            return $tab[$n - 1][$sum] = 0;
    }
}

$n = 5;
$a = array(1, 5, 3, 7, 4);
$sum = 12;

if (subsetSum($a, $n, $sum) != 0) {
    echo "YES\n";
} else {
    echo "NO\n";
}
?>

YES

Time Complexity: O(sum*n)
Auxiliary space: O(n)

PHP Program for Subset Sum Problem using Dynamic Programming:

We can solve the problem in Pseudo-polynomial time we can use the Dynamic programming approach.

So we will create a 2D array of size (n + 1) * (sum + 1) of type boolean. The state dp[i][j] will be true if there exists a subset of elements from set[0 . . . i] with sum value = ‘j’.

The dynamic programming relation is as follows:

if (A[i-1] > j)
dp[i][j] = dp[i-1][j]
else
dp[i][j] = dp[i-1][j] OR dp[i-1][j-set[i-1]]

Below is the implementation of the above approach:

<?php
// A Dynamic Programming solution for 
// subset sum problem

// Returns true if there is a subset of
// set[] with sum equal to given sum
function isSubsetSum( $set, $n, $sum)
{
    // The value of subset[i][j] will
    // be true if there is a subset of
    // set[0..j-1] with sum equal to i
    $subset = array(array());

    // If sum is 0, then answer is true
    for ( $i = 0; $i <= $n; $i++)
        $subset[$i][0] = true;

    // If sum is not 0 and set is empty,
    // then answer is false
    for ( $i = 1; $i <= $sum; $i++)
        $subset[0][$i] = false;

    // Fill the subset table in bottom
    // up manner
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= $sum; $j++)
        {
            if($j < $set[$i-1])
                $subset[$i][$j] = 
                    $subset[$i-1][$j];
            if ($j >= $set[$i-1])
                $subset[$i][$j] = 
                    $subset[$i-1][$j] || 
                    $subset[$i - 1][$j - 
                            $set[$i-1]];
        }
    }

    return $subset[$n][$sum];
}

// Driver code
$set = array(3, 34, 4, 12, 5, 2);
$sum = 9;
$n = count($set);

if (isSubsetSum($set, $n, $sum) == true)
    echo "Found a subset with given sum";
else
    echo "No subset with given sum";

// This code is contributed by anuj_67.
?>

Found a subset with given sum

Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum*n), as the size of the 2-D array is sum*n.

PHP Program for Subset Sum Problem using Dynamic Programming with space optimization to linear:

In previous approach of dynamic programming we have derive the relation between states as given below:
if (A[i-1] > j)
dp[i][j] = dp[i-1][j]
else
dp[i][j] = dp[i-1][j] OR dp[i-1][j-set[i-1]]
If we observe that for calculating current dp[i][j] state we only need previous row dp[i-1][j] or dp[i-1][j-set[i-1]].
There is no need to store all the previous states just one previous state is used to compute result.

Step-by-step approach:

• Define two arrays prev and curr of size Sum+1 to store the just previous row result and current row result respectively.
• Once curr array is calculated then curr becomes our prev for the next row.
• When all rows are processed the answer is stored in prev array.

Below is the implementation of the above approach:

<?php
function isSubsetSum($set, $n, $sum)
{
    $prev = array_fill(0, $sum + 1, false);
    
    for ($i = 0; $i <= $n; $i++)
        $prev[0] = true;

    for ($i = 1; $i <= $sum; $i++)
        $prev[$i] = false;

    $curr = array_fill(0, $sum + 1, false);

    for ($i = 1; $i <= $n; $i++) {
        for ($j = 1; $j <= $sum; $j++) {
            if ($j < $set[$i - 1])
                $curr[$j] = $prev[$j];
            if ($j >= $set[$i - 1])
                $curr[$j] = $prev[$j] || $prev[$j - $set[$i - 1]];
        }
        $prev = $curr;
    }

    return $prev[$sum];
}

$set = array(3, 34, 4, 12, 5, 2);
$sum = 9;
$n = count($set);

if (isSubsetSum($set, $n, $sum)) {
    echo "Found a subset with given sum";
} else {
    echo "No subset with given sum";
}
?>

Found a subset with given sum

Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum), as the size of the 1-D array is sum+1.

Please refer complete article on Subset Sum Problem | DP-25 for more details!

kartik

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