Maximum size subset with given sum
Last Updated :
20 Dec, 2022
This is an extended version of the subset sum problem. Here we need to find the size of the maximum size subset whose sum is equal to the given sum.
Examples:
Input : set[] = {2, 3, 5, 7, 10, 15},
sum = 10
Output : 3
The largest sized subset with sum 10
is {2, 3, 5}
Input : set[] = {1, 2, 3, 4, 5}
sum = 4
Output : 2
This is the further enhancement to the subset sum problem which not only tells whether the subset is possible but also the maximal subset using DP.
To solve the subset sum problem, use the same DP approach as given in the subset sum problem. To further count the maximal subset, we use another DP array (called as 'count array') where count[i][j] is maximal of
- count[i][j-1]. Here current element is not considered.
- count[i- X][j-1] + 1. Here X is the value of the current element selected for the subset.
Implementation:
C++
// A Dynamic Programming solution for subset
// sum problem+ maximal subset value.
#include<bits/stdc++.h>
using namespace std;
// Returns size of maximum sized subset
// if there is a subset of set[] with sun
// equal to given sum. It returns -1 if there
// is no subset with given sum.
int isSubsetSum(int set[], int n, int 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
bool subset[sum + 1][n + 1];
int count[sum + 1][n + 1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
{
subset[0][i] = true;
count[0][i] = 0;
}
// If sum is not 0 and set is empty,
// then answer is false
for (int i = 1; i <= sum; i++)
{
subset[i][0] = false;
count[i][0] = -1;
}
// Fill the subset table in bottom up manner
for (int i = 1; i <= sum; i++)
{
for (int j = 1; j <= n; j++)
{
subset[i][j] = subset[i][j - 1];
count[i][j] = count[i][j - 1];
if (i >= set[j - 1])
{
subset[i][j] = subset[i][j] ||
subset[i - set[j - 1]][j - 1];
if (subset[i][j])
count[i][j] = max(count[i][j - 1],
count[i - set[j - 1]][j - 1] + 1);
}
}
}
return count[sum][n];
}
// Driver code
int main()
{
int set[] = { 2, 3, 5, 10 };
int sum = 20;
int n = 4;
cout<< isSubsetSum(set, n, sum);
}
// This code is contributed by DrRoot_
// A Dynamic Programming solution for subset
// sum problem+ maximal subset value.
import java.util.*;
import java.io.*;
class sumofSub {
// Returns size of maximum sized subset if there
// is a subset of set[] with sun equal to given sum.
// It returns -1 if there is no subset with given sum.
static int isSubsetSum(int set[], int n, int 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
boolean subset[][] = new boolean[sum + 1][n + 1];
int count[][] = new int[sum + 1][n + 1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++) {
subset[0][i] = true;
count[0][i] = 0;
}
// If sum is not 0 and set is empty,
// then answer is false
for (int i = 1; i <= sum; i++) {
subset[i][0] = false;
count[i][0] = -1;
}
// Fill the subset table in bottom up manner
for (int i = 1; i <= sum; i++) {
for (int j = 1; j <= n; j++) {
subset[i][j] = subset[i][j - 1];
count[i][j] = count[i][j - 1];
if (i >= set[j - 1]) {
subset[i][j] = subset[i][j] ||
subset[i - set[j - 1]][j - 1];
if (subset[i][j])
count[i][j] = Math.max(count[i][j - 1],
count[i - set[j - 1]][j - 1] + 1);
}
}
}
return count[sum][n];
}
/* Driver program to test above function */
public static void main(String args[])
{
int set[] = { 2, 3, 5, 10 };
int sum = 20;
int n = set.length;
System.out.println(isSubsetSum(set, n, sum));
}
}
# Python3 program to implement
# the above approach
# A Dynamic Programming solution
# for subset sum problem+ maximal
# subset value.
# Returns size of maximum sized subset
# if there is a subset of set[] with sun
# equal to given sum. It returns -1 if there
# is no subset with given sum.
def isSubsetSum(arr, 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 = [[False for x in range(n + 1)]
for y in range (sum + 1)]
count = [[0 for x in range (n + 1)]
for y in range (sum + 1)]
# If sum is 0, then answer is true
for i in range (n + 1):
subset[0][i] = True
count[0][i] = 0
# If sum is not 0 and set is empty,
# then answer is false
for i in range (1, sum + 1):
subset[i][0] = False
count[i][0] = -1
# Fill the subset table in bottom up manner
for i in range (1, sum + 1):
for j in range (1, n + 1):
subset[i][j] = subset[i][j - 1]
count[i][j] = count[i][j - 1]
if (i >= arr[j - 1]) :
subset[i][j] = (subset[i][j] or
subset[i - arr[j - 1]][j - 1])
if (subset[i][j]):
count[i][j] = (max(count[i][j - 1],
count[i - arr[j - 1]][j - 1] + 1))
return count[sum][n]
# Driver code
if __name__ == "__main__":
arr = [2, 3, 5, 10]
sum = 20
n = 4
print (isSubsetSum(arr, n, sum))
# This code is contributed by Chitranayal
// A Dynamic Programming solution for subset
// sum problem+ maximal subset value.
using System;
class sumofSub {
// Returns size of maximum sized subset
// if there is a subset of set[] with sun
// equal to given sum. It returns -1 if there
// is no subset with given sum.
static int isSubsetSum(int[] set, int n, int 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
bool[, ] subset = new bool[sum + 1, n + 1];
int[, ] count = new int[sum + 1, n + 1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++) {
subset[0, i] = true;
count[0, i] = 0;
}
// If sum is not 0 and set is empty,
// then answer is false
for (int i = 1; i <= sum; i++) {
subset[i, 0] = false;
count[i, 0] = -1;
}
// Fill the subset table in bottom up manner
for (int i = 1; i <= sum; i++) {
for (int j = 1; j <= n; j++) {
subset[i, j] = subset[i, j - 1];
count[i, j] = count[i, j - 1];
if (i >= set[j - 1]) {
subset[i, j] = subset[i, j] ||
subset[i - set[j - 1], j - 1];
if (subset[i, j])
count[i, j] = Math.Max(count[i, j - 1],
count[i - set[j - 1], j - 1] + 1);
}
}
}
return count[sum, n];
}
/* Driver program to test above function */
public static void Main()
{
int[] set = { 2, 3, 5, 10 };
int sum = 20;
int n = set.Length;
Console.WriteLine(isSubsetSum(set, n, sum));
}
}
// This code is contributed by vt_m.
<script>
// JavaScript Programming solution for subset
// sum problem+ maximal subset value.
// Returns size of maximum sized subset
// if there is a subset of set[] with
// sun equal to given sum. It returns -1
// if there is no subset with 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
let subset = new Array(sum + 1);
// Loop to create 2D array using 1D array
for(var i = 0; i < subset.length; i++)
{
subset[i] = new Array(2);
}
let count = new Array(sum + 1);
// Loop to create 2D array using 1D array
for(var i = 0; i < count.length; i++)
{
count[i] = new Array(2);
}
// If sum is 0, then answer is true
for(let i = 0; i <= n; i++)
{
subset[0][i] = true;
count[0][i] = 0;
}
// If sum is not 0 and set is empty,
// then answer is false
for(let i = 1; i <= sum; i++)
{
subset[i][0] = false;
count[i][0] = -1;
}
// Fill the subset table in bottom up manner
for(let i = 1; i <= sum; i++)
{
for(let j = 1; j <= n; j++)
{
subset[i][j] = subset[i][j - 1];
count[i][j] = count[i][j - 1];
if (i >= set[j - 1])
{
subset[i][j] = subset[i][j] ||
subset[i - set[j - 1]][j - 1];
if (subset[i][j])
count[i][j] = Math.max(count[i][j - 1],
count[i - set[j - 1]][j - 1] + 1);
}
}
}
return count[sum][n];
}
// Driver Code
let set = [ 2, 3, 5, 10 ];
let sum = 20;
let n = set.length;
document.write(isSubsetSum(set, n, sum));
// This code is contributed by sanjoy_62
</script>
Time complexity: O(sum*n).
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