Added for 19 Jan

GeeksForGeeks/19-1-24/GFG.py

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+#User function Template for python3
+
+class Solution:
+    def kTop(self, arr, N, K):
+        final_ans=[]
+        top = [0 for i in range(K + 1)] 
+        freq = {i:0 for i in range(K + 1)} 
+
+        for m in range(N): 
+            if a[m] in freq.keys(): 
+                freq[a[m]] += 1
+            else: 
+                freq[a[m]] = 1
+
+            top[k] = a[m] 
+            i = top.index(a[m]) 
+            i -= 1
+
+            while i >= 0: 
+                if (freq[top[i]] < freq[top[i + 1]]): 
+                    t = top[i] 
+                    top[i] = top[i + 1] 
+                    top[i + 1] = t 
+
+                elif ((freq[top[i]] == freq[top[i + 1]]) and (top[i] > top[i + 1])): 
+                    t = top[i] 
+                    top[i] = top[i + 1] 
+                    top[i + 1] = t 
+                else: 
+                    break
+                i -= 1
+
+            i = 0
+            ans = []
+            while i < K and top[i] != 0: 
+                ans.append(top[i]) 
+                i += 1
+
+            final_ans += [ans]
+        return final_ans
+
+#{ 
+ # Driver Code Starts
+
+
+t=int(input())
+for _ in range(0,t):
+    n,k=list(map(int,input().split()))
+    a=list(map(int,input().split()))
+    ob = Solution()
+    ans=ob.kTop(a,n,k)
+    for line in ans:
+        print(*line)
+
+
+
+# } Driver Code Ends

GeeksForGeeks/19-1-24/README.md

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+Time complexity - O(n * k)
+Space complexity - O(n)
+
+Was not possible for me to return the output as said in Java (ArrayList<Integer>); could had done if it was 2d ArrayList

LeetCode/19-1-24/README.md

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+Time complexity - O(n^2)
+Space complexity - O(1)

LeetCode/19-1-24/Solution.java

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+class Solution 
+{
+    public int minFallingPathSum(int[][] A) 
+    {
+        final int n = A.length;
+
+        for (int i = 1; i < n; ++i)
+            for (int j = 0; j < n; ++j) 
+            {
+                int min = Integer.MAX_VALUE;
+
+                for (int k = Math.max(0, j - 1); k < Math.min(n, j + 2); ++k)
+                    min = Math.min(min, A[i - 1][k]);
+
+                A[i][j] += min;
+            }
+
+        return Arrays.stream(A[n - 1]).min().getAsInt();
+    }
+}