Merge pull request rachitiitr#52 from rachitiitr/youtube_live_dp

LIVE_YOUTUBE - 3 DP problems
YouTube live stream: https://youtu.be/mxbFPfHMX3s
First problem: https://leetcode.com/problems/maximum-subarray
Second problem: https://leetcode.com/problems/matrix-block-sum/
Third problem: https://leetcode.com/problems/reducing-dishes/

Author: rachitiitr

LeetCode/1314.matrix-block-sum.cpp

@@ -0,0 +1,69 @@
+class Solution {
+public:
+    vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int k) {
+        int n = mat.size(), m = mat[0].size();
+        vector<vector<int>> ans(n, vector<int>(m, 0));
+        vector<vector<int>> dp(n+1, vector<int>(m+1, 0)); // 1-based indexing <-- prefix sums,˘
+        
+        // compute the prefix sums
+        for(int i = 1; i <= n; i++) {
+            for(int j = 1; j <= m; j++) {
+                dp[i][j] = mat[i-1][j-1] + dp[i][j-1] + dp[i-1][j] - dp[i-1][j-1];
+            }
+        }
+        
+        for(int i = 0; i < n; i++) {
+            for(int j = 0; j < m; j++) {
+                // compute the ans for (i,j)
+                int i1 = max(0, i-k), j1 = max(0,j-k);
+                int i2 = min(n-1, i+k), j2 = min(m-1, j+k);
+                i1++, i2++, j1++, j2++; //1-based indexing
+               
+                ans[i][j] = dp[i2][j2] - dp[i2][j1-1] - dp[i1-1][j2] + dp[i1-1][j1-1];
+            }
+        },™
+        return ans;
+    }
+};
+
+/*
+prefix sum in 2d approach
+dp[i][j] = sum of rect from (0,0) to (i,j) as diagnal points
+=> it helps me in computing any rectangle sum in O(1)
+(i1, j1) and (i2, j2) i1 < i2 and j1 < j2
+
+dp[i2][j2] - dp[i1][j1-1] - dp[i1-1][j2] + dp[i1-1][j1-1]
+
+
+
+prefix sums can be used for each row individually
+=> for given row, sum(j-k, j+k) => O(1)
+      j-k       j       j+k
+                 
+                 
+i-k             
+                 
+                 
+                 
+i               x
+
+
+
+i+k
+
+(i,j) => O(k^2)
+O(N^2k^2)
+O(N^2 k)
+          
+          
+          
+
+k = 1
+[1,2,3],
+[4,5,6],
+[7,8,9]
+
+9*10/2 = 45
+    
+    5+6+8+9 = 28
+*/

LeetCode/1402.reducing-dishes.cpp

@@ -0,0 +1,29 @@
+class Solution {
+public:
+    int maxSatisfaction(vector<int>& arr) {
+        int n = arr.size();
+        sort(arr.begin(), arr.end());
+        int ans = 0;
+        for(int i = 0; i < n; i++) {
+            int cur = 0;
+            // considering the suffix from pos i
+            // [arr[i], arr[i+1], ... arr[n-1]]
+               // 1         2              
+            for(int j = i; j < n; j++) {
+                cur += (j-i+1) * arr[j];
+            }
+            ans = max(ans, cur);
+        }
+        return ans;
+    }
+};
+
+/*
+a1 <= a2 <= a3 .... // we have proved
+(a1,a2,...,ak) <-- reordering of satisfaction array + you can remove elements
+max(a1 + 2*a2 + 3*a3 + k*ak)
+    
+    -9   -8  -1   0   5
+    
+    how long a prefix would you remove to get the max score
+*/

LeetCode/53.maximum-subarray.cpp

@@ -0,0 +1,27 @@
+class Solution {
+public:
+    int maxSubArray(vector<int>& nums) {
+        int n = nums.size();
+        int ans = INT_MIN;
+        
+        // vector<int> dp(n+1, 0); // O(N) extra space
+        // dp[i] = max subarray sum starting from index i
+        // mandate to pick something
+        int nextMax = 0;
+        for(int i = n-1; i >= 0; i--) {
+            // dp[i] = max(nums[i] + dp[i+1], 0)
+            // dp[i] = nums[i] + max(dp[i+1], 0);
+            int curMax = nums[i] + max(nextMax, 0);
+            ans = max(ans, curMax);
+            nextMax = curMax;
+        }
+       
+        return ans;
+    }
+};
+
+/*
+_ _ _ _ _ 100 _ _ _ _ _ 
+        i ^
+        dp[i] = max(nums[i] + dp[i+1], 0)
+*/