diff --git a/Amortized1/README.md b/Amortized1/README.md
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+# Solution Approach
+
+Note that the variable j is not initialized for each value of variable i.
+Hence, the inner j++ will be executed at most n times.
+The i loop also runs n times.
+So, the whole thing runs for O(n) times.
+
+
+**Still not convinced ?**
+
+Lets assume the array passed has its element in decreasing order. We will just dry run through the code :
+
+Iteration 1 : i = 0, j = 0. arr[0] < arr[0] is false. So, the inner while loop breaks.
+Iteration 2: i =1, j = 0. arr[1] < arr[0] is true. j becomes 1.
+Iteration 3 : i = 1, j = 1. Condition false. We break. Note that j will remain 1 and is not reset back to 0.
+Iteration 4 : i = 2, j = 1. arr[2] < arr[1]. True. j = 2.
+Iteration 5 : i = 2, j = 2. Condition false. Break.
+Iteration 6 : i = 3, j = 2. arr[3] < arr[2]. True. j = 3.
+Iteration 7 : i = 3, j = 3. Condition false. Break.
+
+
+As you can see, the inner while loop only runs once in this case.
+So, total iterations is **2 * N.**
diff --git a/Choose1/README.md b/Choose1/README.md
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+# Solution Approach
+
+The order of growth of option (C) is n^2.5 which is higher than n^2.
+
+Lets look at it with a different approach :
+
+f(n) = O(n^2) if
+f(n) <= C * n^2 for n > n0.
+
+
+Lets look at every option one by one.
+
+* Option 1 :
+
+(15^10) * n + 12099
+Lets say C = 16^10
+15^10 * n + 12099 < 16^10 * n^2 for n > 1.
+So, it is O(n^2).
+
+
+* Option 2 : n^1.98
+
+C = 1.
+    n^1.98 < n^2 for n > 1.
+So, its O(n^2) ```
+Option 3 : n^3 / (sqrt(n)) or n^2.5
+    There is no possible combination of C and n0 possible
+Option 4 : 2^20 * n
+    C = 2^20, n0 = 1
+        2^20 * n <= 2^20 * n^2 for n > 1
diff --git a/Choose2/README.md b/Choose2/README.md
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+++ b/Choose2/README.md
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+# Solution Approach
+
+Try to look at the values for functions for very large value of n.
diff --git a/Choose3/README.md b/Choose3/README.md
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+# Solution Approach
+
+The time complexity of the first for loop is O(n). 
+
+The time complexity of the second for loop is O(n/2), equivalent to O(n) in asymptotic analysis. 
+
+The time complexity of the third for loop is O(logn). 
+
+The fourth for loop doesn't terminate. 
diff --git a/Choose4/README.md b/Choose4/README.md
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+++ b/Choose4/README.md
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+# Solution Approach
+
+In asymptotic analysis we consider growth of algorithm in terms of input size.
+An algorithm X is said to be asymptotically better than Y if X takes smaller time than y for all input sizes n larger than a value n0 where n0 > 0.
diff --git a/GcdCmpl/README.md b/GcdCmpl/README.md
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+# Solution Approach
+
+Let us say n = fibonacci(N) and m = fibonacci(N - 1)
+
+fibonacci(N) = fibonacci(N-1) + fibonacci(N-2)
+
+OR n = m + k where k < m. 
+
+Therefore the step 
+
+    n = n % m will make n = k
+    
+    swap(n, m) will result in
+    
+    n = fibonacci(N-1)
+        
+    m = k = fibonacci(N-2)
+    
+So, it will take N steps before m becomes 0.
+
+This means, in the worst case, this algorithm can take N step if **n** is Nth fibonacci number. 
+
+**Think of whats the relation between N and n**. 
diff --git a/LOOP_CMPL/README.md b/LOOP_CMPL/README.md
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+++ b/LOOP_CMPL/README.md
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+# solution approach
+
+The first loop is O(N) and the second loop is O(M). 
+
+Since you don't know which is bigger, you say **this is O(N + M)**. This can also be written as O(max(N, M)).
+
+Since there is no additional space being utilised, the space complexity is constant / O(1)
diff --git a/LoopCmpl2/README.md b/LoopCmpl2/README.md
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+# Solution Approach
+
+Lets look at the code we are evaluating : 
+
+int i, j, k = 0;
+for (i = n/2; i <= n; i++) {
+for (j = 2; j <= n; j = j * 2) {
+k = k + n/2;
+}
+}
+
+
+Now, lets just assume `n = 8` for now. 
+We will try to see, the values of j corresponding to each i. 
+
+i = 4, j = 2, 4, 8
+i = 5, j = 2, 4, 8
+i = 6, j = 2, 4, 8
+i = 7, j = 2, 4, 8
+i = 8, j = 2, 4, 8
+
+
+If you notice, j keeps doubling till it is less than or equal to n. Number of times, you can double a number till it is less than n would be log(n). 
+
+Lets take more examples here to convince ourselves.
+
+n = 16, j = 2, 4, 8, 16
+n = 32, j = 2, 4, 8, 16, 32
+
+
+So, j would run for O(log n) steps. 
+i runs for n/2 steps. 
+
+So, total steps ` = O (n/ 2 * log (n)) = O(n logn) `
diff --git a/NestedCmpl/README.md b/NestedCmpl/README.md
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+# Solution Approach
+
+The first set of nested loops is O(N^2) and the second loop is O(N). 
+
+This is O(max(N^2,N)) which is O(N^2). 
diff --git a/NestedCmpl2/README.md b/NestedCmpl2/README.md
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+# Solution Approach
+
+Total number of runs = N + (N - 1) + (N - 2) + ... 1 + 0
+
+= N * (N + 1) / 2
+
+= 1/2 * N^2 + 1/2 * N
+
+O(N^2) times. 
diff --git a/NestedCmpl3/README.md b/NestedCmpl3/README.md
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+# Solution Approach
+
+In the first iteration, the j loop runs N times.
+
+In the second iteration, the j loop runs N / 2 times. 
+
+In the ith iteration, the j loop runs N / 2^i times. 
+
+So, the total number of runs of loop = N + N / 2 + N / 4 + ... 1 
+
+= **N * ( 1 + 1/2 + 1/4 + 1/8 + ... ) < 2 * N** 
diff --git a/RecCmpl1/README.md b/RecCmpl1/README.md
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+# Solution Approach
+
+It might seem at the first look that the program is O(log N). 
+However, the last case 
+
+return searchNumOccurrence(V, k, start, mid - 1) + 1 + searchNumOccurrence(V, k, mid + 1, end);
+
+
+is the bottleneck step. 
+Assuming all the values in the array are the same, we get the following relation : 
+
+T(N) = 2 * T(N/2) + constant
+= 4 * T(N/4) + constant ( 2 * constant = another constant )
+= 8 * T(N/8) + constant
+…
+= N * T(N/N) + constant
+= N + constant
+= O(N)
diff --git a/RecCmpl2/README.md b/RecCmpl2/README.md
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+# Solution Approach
+
+Note that in every function call, we end up making 2 calls.
+
+So, the function calls ends up looking like a tree :
+
+                   F(0,0)
+                /          \ 
+            F(0, 1)         F(1, 0)
+             /    \         /       \ 
+          F(0, 2)  F(1,1)  F(1, 1)  F(2, 0)
+             ....
+The function calls end up making a complete binary tree.
+
+  Number of calls on Level 0 = 1
+  Number of calls on Level 1 = 2
+  Number of calls on Level 2 = 4
+  ...
+  Number of calls on level i = 2^i. 
+Total number of calls = 1 + 2 + 4 + ... 2^i + ... 2^(M + N - 2)  
+                      = O(2^(M + N))
diff --git a/RecCmpl3/README.md b/RecCmpl3/README.md
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+# Solution Approach
+
+Note that for a given `(r, c)`
+ 
+the following code will not be executed more than once : 
+
+memo[r][c] = V[r][c] + min(findMinPath(V, r + 1, c), findMinPath(V, r, c + 1));
+
+
+Once memo[r][c] is set, the functions will return at 
+
+if (memo[r][c] != -1) return memo[r][c];
+
+
+So, every function ends up calling other functions at most 1 time. 
+In other words, every function ends up executing atmost O(1) times (Note that you can shift the part about checking for memo[r][c] != -1 at the callsite ).
+
+`O(R * C)` possible number of combinations are possible for `(r, c)`
+Hence, the time complexity of the function : O(R*C)
diff --git a/WhileCmpl/README.md b/WhileCmpl/README.md
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+++ b/WhileCmpl/README.md
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+# Solution Approach
+
+We have to find the smallest x such that `N / 2^x < 1 OR 2^x > N`
+
+x = log(N)