Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/24 - Partition Equal Subset Sum/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+public:
+    // Function to check if a subset sum equal to 'target' exists using dynamic programming with space optimization
+    int solve(vector<int>& nums, int n, int target) {
+        // Arrays to store the current and next row of the dp table
+        vector<int> curr(target+1, 0);   // To store the result for the current index
+        vector<int> next(target+1, 0);   // To store the result for the next index (used in the transition)
+
+        // Base case: sum of 0 can always be formed (empty subset)
+        curr[0] = 1;
+        next[0] = 1;
+
+        // Iterate over all indices of nums in reverse (from n-1 to 0)
+        for(int index = n-1; index >= 0; index--) {
+            // Iterate over all possible target sums from 1 to the desired target sum
+            for(int t = 1; t <= target; t++) {
+                int include = 0;
+                // If the current number can be included (i.e., target - nums[index] >= 0)
+                if(t - nums[index] >= 0) include = next[t - nums[index]];
+
+                // Exclude the current element (i.e., use the result from the next index with the same target)
+                int exclude = next[t];
+
+                // The result for curr[t] is the logical OR of including or excluding the element
+                curr[t] = include or exclude;
+            }
+
+            // Move the current row to the next row for the next iteration
+            next = curr;
+        }
+
+        // The final result is stored in next[target], which tells if the target sum can be achieved
+        return next[target];
+    }
+
+    // Main function to check if we can partition the array into two subsets with equal sum
+    bool canPartition(vector<int>& nums) {
+        int n = nums.size();
+        int total = 0;
+
+        // Calculate the total sum of the elements in the array
+        for(auto & num : nums) total += num;
+
+        // If the total sum is odd, it can't be partitioned into two equal subsets
+        if(total & 1) return 0;
+
+        // Set the target as half of the total sum
+        int target = total / 2;
+
+        // Call the solve function to check if a subset with sum equal to 'target' exists
+        return solve(nums, n, target);
+    }
+};