Added Solution to both the problems

Problem1.java

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+// Time Complexity : O(n + maxValue)
+// Space Complexity : O(maxValue)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+class Solution {
+    public int deleteAndEarn(int[] nums) {
+        // Edge case
+        if (nums.length== 0) return 0;
+        int max=0;
+        for (int n:nums) {
+            max = Math.max(max, n);
+        }
+        // points[i] = total points we get if we take number i
+        int[] points = new int[max + 1];
+        for (int n : nums) {
+            points[n]=points[n]+n;
+        }
+        // Same as house robber
+        int prev2 = 0;// dp[i-2]
+        int prev1 = points[0]; // dp[i-1]
+        for (int i=1;i<= max;i++) {
+            int curr = Math.max(prev1,prev2+points[i]);
+            prev2=prev1;
+            prev1=curr;
+        }
+        return prev1;
+    }
+}
+

Problem2.java

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+// Time Complexity : O(n*n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+class Solution {
+    public int minFallingPathSum(int[][] matrix) {
+        int n = matrix.length;
+        // Start from second row
+        for (int i=1;i<n;i++) {
+            for (int j=0;j<n;j++) {
+                // down(i+1,j)
+                // down-left(i+1,j-1)
+                // down-right(i+1,j+1)
+                int up=matrix[i-1][j];
+                int left=(j>0)?matrix[i-1][j-1]:Integer.MAX_VALUE;
+                int right =(j<n-1)?matrix[i-1][j+1]: Integer.MAX_VALUE;
+                // Add minimum of possible paths
+                matrix[i][j] += Math.min(up, Math.min(left, right));
+            }
+        }
+        // Answer is minimum in last row
+        int ans = Integer.MAX_VALUE;
+        for (int j=0; j<n;j++) {
+            ans=Math.min(ans,matrix[n - 1][j]);
+        }
+        return ans;
+    }
+}
+