Create 05 - Space Optimized - 2 | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/19 - Longest Increasing Subsequence/05 - Space Optimized - 2 | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to calculate the length of the Longest Increasing Subsequence (LIS)
+    // using a simple dynamic programming approach.
+    int solve(vector<int>& nums) {
+        int n = nums.size(); // Get the size of the array.
+
+        // Create a 1D DP array where dp[i] represents the length of the LIS ending at index `i`.
+        vector<int> dp(n, 1);
+
+        // Iterate through the array to build the DP table.
+        for (int currIndex = 1; currIndex < n; currIndex++) {
+            // For each `currIndex`, check all previous indices (`prevIndex`) to find
+            // the longest increasing subsequence that can be extended by `nums[currIndex]`.
+            for (int prevIndex = 0; prevIndex < currIndex; prevIndex++) {
+                // If the current element is greater than the previous element,
+                // update dp[currIndex] to include the LIS ending at `prevIndex`.
+                if (nums[currIndex] > nums[prevIndex]) {
+                    dp[currIndex] = max(dp[currIndex], dp[prevIndex] + 1);
+                }
+            }
+        }
+
+        // The length of the LIS is the maximum value in the DP array.
+        // Use `max_element` to find the maximum value in the DP array.
+        return *max_element(dp.begin(), dp.end());
+    }
+
+    // Main function to calculate the length of the Longest Increasing Subsequence (LIS).
+    int lengthOfLIS(vector<int>& nums) {
+        int n = nums.size(); // Get the size of the array.
+        return solve(nums); // Call the helper function to solve the problem.
+    }
+};