Subset sum

shriyanshi24jindal

Mar 28th, 2026

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Java 21.86 KB | None | 0 0

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class Solution {
/** Recursive code */
static Boolean isSubsetSum(int arr[], int sum) {
int n = arr.length;
return findSubsetSum(arr, sum, n);
}
static Boolean findSubsetSum(int[] arr, int sum, int n)
{
// when we reach end of the array, so couldn't found a subset
if(n == 0)
return false;
// when sum becomes 0 that means found a subset
if(sum == 0)
return true;
if(arr[n-1] <= sum)
{
// pick the element in subset
boolean pick = findSubsetSum(arr, sum - arr[n-1], n-1);
// do not pick the element in subset
boolean notPick = findSubsetSum(arr, sum, n-1);
return pick || notPick;
}
// when current element exceeds given sum, so skip the element
return findSubsetSum(arr, sum, n-1);
}
/** Memoization code */
static Boolean isSubsetSum(int arr[], int sum) {
int n = arr.length;
boolean[][] t = new boolean[n+1][sum+1];
// when array of size will increase but sum needs to always 0, then we can have always empty subset
for(int i=0; i<n+1; i++)
{
Arrays.fill(t[i], true);
}
// when sum increases but size of array is always 0, then we cannot have any subset
for(int j=0; j<n+1; j++)
{
Arrays.fill(t[j], false);
}
return findSubsetSum(arr, sum, n, t);
}
static Boolean findSubsetSum(int[] arr, int sum, int n, boolean[][] t)
{
// when we reach end of the array, so couldn't found a subset
if(n == 0)
return false;
// when sum becomes 0 that means found a subset
if(sum == 0)
return true;
if(t[n][sum] != false)
return t[n][sum];
if(arr[n-1] <= sum)
{
// pick the element in subset
boolean pick = findSubsetSum(arr, sum - arr[n-1], n-1, t);
// do not pick the element in subset
boolean notPick = findSubsetSum(arr, sum, n-1, t);
t[n][sum] = pick || notPick;
return t[n][sum];
}
// when current element exceeds given sum, so skip the element
t[n][sum] = findSubsetSum(arr, sum, n-1, t);
return t[n][sum];
}
/** Top-down approach */
static Boolean isSubsetSum(int arr[], int sum) {
int n = arr.length;
boolean[][] t = new boolean[n+1][sum+1];
// intialise the matrix on basis of base condition
for(int i=0; i<n+1; i++)
{
for(int j=0; j<sum+1; j++)
{
// when sum increases but size of array is always 0, then we cannot have any subset
if(i==0)
{
t[i][j] = false;
}
// when array of size will increase but sum needs to always 0, then we can have always empty subset
if(j==0)
{
t[i][j] = true;
}
}
}
for(int i=1; i<n+1; i++)
{
for(int j=1; j<sum+1; j++)
{
if(arr[i-1] <= j)
{
boolean pick = t[i-1][j - arr[i-1]];
boolean notPick = t[i-1][j];
t[i][j] = pick || notPick;
}
else
t[i][j] = t[i-1][j];
}
}
return t[n][sum];
}
}

RAW Paste Data Copied

class Solution {
    
    /** Recursive code */
    static Boolean isSubsetSum(int arr[], int sum) {
        int n = arr.length;
        return findSubsetSum(arr, sum, n);
    }
    
    static Boolean findSubsetSum(int[] arr, int sum, int n)
    {
        // when we reach end of the array, so couldn't found a subset
        if(n == 0)
            return false;
        
        // when sum becomes 0 that means found a subset
        if(sum == 0)
            return true;
            
        if(arr[n-1] <= sum)
        {
            // pick the element in subset
            boolean pick = findSubsetSum(arr, sum - arr[n-1], n-1);
            // do not pick the element in subset
            boolean notPick = findSubsetSum(arr, sum, n-1);
            return pick || notPick;
        }
        // when current element exceeds given sum, so skip the element
        return findSubsetSum(arr, sum, n-1);
    }
    
    /** Memoization code */
    static Boolean isSubsetSum(int arr[], int sum) {
        int n = arr.length;
        
        boolean[][] t = new boolean[n+1][sum+1];
        // when array of size will increase but sum needs to always 0, then we can have always empty subset
        for(int i=0; i<n+1; i++)
        {
            Arrays.fill(t[i], true);
        }
        
        // when sum increases but size of array is always 0, then we cannot have any subset
        for(int j=0; j<n+1; j++)
        {
            Arrays.fill(t[j], false);
        }
        
        return findSubsetSum(arr, sum, n, t);
    }
    
    static Boolean findSubsetSum(int[] arr, int sum, int n, boolean[][] t)
    {
        // when we reach end of the array, so couldn't found a subset
        if(n == 0)
            return false;
        
        // when sum becomes 0 that means found a subset
        if(sum == 0)
            return true;
            
        if(t[n][sum] != false)
            return t[n][sum];
            
        if(arr[n-1] <= sum)
        {
            // pick the element in subset
            boolean pick = findSubsetSum(arr, sum - arr[n-1], n-1, t);
            // do not pick the element in subset
            boolean notPick = findSubsetSum(arr, sum, n-1, t);
            t[n][sum] = pick || notPick;
            return t[n][sum];
        }
        // when current element exceeds given sum, so skip the element
        t[n][sum] = findSubsetSum(arr, sum, n-1, t);
        return t[n][sum];
    }
    
    /** Top-down approach */
    static Boolean isSubsetSum(int arr[], int sum) {
        int n = arr.length;
        
        boolean[][] t = new boolean[n+1][sum+1];
        
        // intialise the matrix on basis of base condition
        for(int i=0; i<n+1; i++)
        {
            for(int j=0; j<sum+1; j++)
            {
                // when sum increases but size of array is always 0, then we cannot have any subset
                if(i==0)
                {
                    t[i][j] = false;
                }
                // when array of size will increase but sum needs to always 0, then we can have always empty subset
                if(j==0)
                {
                     t[i][j] = true;
                }
            }
        }
        
        for(int i=1; i<n+1; i++)
        {
            for(int j=1; j<sum+1; j++)
            {
                if(arr[i-1] <= j)
                {
                    boolean pick = t[i-1][j - arr[i-1]];
                    boolean notPick = t[i-1][j];
                    t[i][j] = pick || notPick;
                }
                else
                    t[i][j] = t[i-1][j];
            }
        }
        return t[n][sum];
    }
}