PROBLEM DESCRIPTION
Given an array arr[] of size N, check if it can be partitioned into two parts such that the sum of elements in both parts is the same.
SOLUTION
BRUTE-FORCE (TLE)
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class Solution{
static int equalPartition(int n, int arr[])
{
int sum = 0;
for(int i=0; i<n; i++)
sum += arr[i];
if(sum%2 == 1)
return 0;
int halfSum = sum/2;
return partitionHelper(n, arr, 0, halfSum, 0) ? 1 : 0;
}
static boolean partitionHelper(int n, int[] arr, int currentSum, int halfSum, int idx){
if(idx == n){
return currentSum == halfSum;
}
// include
boolean include = partitionHelper(n, arr, currentSum + arr[idx], halfSum, idx+1);
// exclude
boolean exclude = partitionHelper(n, arr, currentSum, halfSum, idx+1);
return include || exclude;
}
}
DYNAMIC PROGRAMMING (ACCEPTED)
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class Solution{
static int equalPartition(int n, int arr[])
{
// Calculate the total sum of the array
int sum = 0;
for(int i = 0; i < n; i++)
sum += arr[i];
// If the total sum is odd, partitioning into two equal subsets is not possible
if(sum % 2 == 1)
return 0;
// The target sum for each partition will be half of the total sum
int halfSum = sum / 2;
// dp[i][j]: if it's possible to form sum j using the first i elements of the array
boolean[][] dp = new boolean[n][halfSum + 1];
// It's always possible to form a sum of 0 by using 0 elements
for(int i = 0; i < n; i++)
dp[i][0] = true;
// Loop through the elements
for(int i = 1; i < n; i++){
// Is sum j possible using first i elements
for(int j = 1; j <= halfSum; j++){
// Option 1: Exclude the current element from the subset
dp[i][j] = dp[i - 1][j];
// Option 2: Include the current element in the subset, if it doesn't exceed the current sum
if (j >= arr[i])
// Check if sum j-arr[i] can be formed with previous elements
dp[i][j] |= dp[i - 1][j - arr[i]];
}
}
// If dp[n-1][halfSum] is true, it's possible to partition the array into two equal subsets
return dp[n-1][halfSum] ? 1 : 0;
}
}
ANOTHER APPROACH
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class Solution{
static int equalPartition(int n, int arr[])
{
// Calculate the total sum of the array
int sum = 0;
for(int i = 0; i < n; i++)
sum += arr[i];
// If the total sum is odd, we cannot split it into two equal subsets
if(sum % 2 == 1)
return 0;
// The target sum for each subset is half of the total sum
int halfSum = sum / 2;
// Use a HashSet to store all possible subset sums, starting with 0 (sum of an empty set)
Set<Integer> sumList = new HashSet<>();
sumList.add(0);
// Iterate through each element in the array
for(int i = 0; i < n; i++){
// Create a new set to store possible sums by adding the current element
Set<Integer> currentSumList = new HashSet<>();
// For each sum already in sumList, add the current element and store the new sum
for(Integer x : sumList)
currentSumList.add(arr[i] + x);
// Add all the new sums to the main set
sumList.addAll(currentSumList);
}
// Check if halfSum is in sumList, meaning we can partition the array into two equal subsets
return sumList.contains(halfSum) ? 1 : 0;
}
}