Added LeetCode and Geeks for Geeks Solutions in CPP

C++/Geeks4Geeks/1's Complement/1s-complement.cpp

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+// { Driver Code Starts
+
+#include<bits/stdc++.h>
+using namespace std; 
+
+ // } Driver Code Ends
+
+class Solution{ 
+public:
+    string onesComplement(string S,int N){
+        //code here
+        for(int i=0;i<N;i++){
+            if(S[i] == '0'){
+                S[i] = '1';
+            }
+            else{
+                S[i] = '0';
+            }
+        }
+    return S;
+    }
+};
+
+// { Driver Code Starts.
+int main()
+{
+    int t;
+    cin>>t;
+    while(t--)
+    {
+        int n;
+        cin>>n;
+        string s;
+        cin>>s;
+        Solution ob;
+        cout<<ob.onesComplement(s,n)<<"\n";
+    }
+}  // } Driver Code Ends

C++/Geeks4Geeks/1's Complement/README.md

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+# 1's Complement
+## Basic 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given an<strong> N </strong>bit binary number, find the 1's complement of the number.&nbsp;The ones'&nbsp;complement&nbsp;of a binary number is&nbsp;defined&nbsp;as the value obtained by inverting all the bits in the binary representation of the number (swapping 0s for 1s and vice versa).</span><br>
+&nbsp;</p>
+
+<p><strong><span style="font-size:18px">Example 1:</span></strong></p>
+
+<pre><strong><span style="font-size:18px">Input:</span>
+</strong><span style="font-size:18px">N = 3
+S = 101
+<strong>Output:
+</strong>010
+<strong>Explanation:
+</strong>We get the output by converting 1's in S
+to 0 and 0s to 1</span><strong>
+</strong></pre>
+
+<p><strong><span style="font-size:18px">Example 2:</span></strong></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong>
+N = 2
+S = 10
+<strong>Output:</strong>
+01
+<strong>Explanation:</strong>
+We get the output by converting 1's in S
+to 0 and 0s to 1</span>
+</pre>
+
+<p><br>
+<span style="font-size:18px"><strong>Your Task:&nbsp;&nbsp;</strong><br>
+You don't need to read input or print anything. Your task is to complete the function&nbsp;<strong>onesComplement()</strong>&nbsp;which takes the binary string S, its size N<strong>&nbsp;</strong>as input parameters&nbsp;and returns 1's complement of S of size N.</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Expected Time Complexity:</strong> O(N)<br>
+<strong>Expected Space Complexity:</strong> O(N)</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+1&lt;=N&lt;=100</span></p>
+ <p></p>
+            </div>

C++/Geeks4Geeks/Binary Search/README.md

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+# Binary Search
+## Basic 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given a sorted array of size N and an integer K, find the position at which K is present in the array using binary search.</span></p>
+
+<p><br>
+<span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong>
+N = 5
+arr[] = {1 2 3 4 5} 
+K = 4
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> 4 appears at index 3.</span></pre>
+
+<p><br>
+<span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong>
+N = 5
+arr[] = {11 22 33 44 55} 
+K = 445
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> 445 is not present.</span></pre>
+
+<p><br>
+<span style="font-size:18px"><strong>Your Task: &nbsp;</strong><br>
+You dont need to read input or print anything. Complete the function <strong>binarysearch()</strong> which takes arr[], N and K as input parameters and returns the index of K in the array. If K is not present in the array, return -1.</span></p>
+
+<p><br>
+<span style="font-size:18px"><strong>Expected Time Complexity:</strong> O(LogN)<br>
+<strong>Expected Auxiliary Space:</strong> O(LogN) if solving recursively and O(1) otherwise.</span></p>
+
+<p><br>
+<span style="font-size:18px"><strong>Constraints:</strong><br>
+1 &lt;= N &lt;= 10<sup>4</sup><br>
+1 &lt;= arr[i] &lt;= 10<sup>4</sup></span></p>
+ <p></p>
+            </div>

C++/Geeks4Geeks/Binary number to decimal number/README.md

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+# Binary number to decimal number
+## Basic 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given a Binary Number <strong>B</strong>, find&nbsp;its decimal equivalent.</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input: </strong>B = 10001000
+<strong>Output: </strong>136</span>
+</pre>
+
+<p><span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input: </strong>B = 101100
+<strong>Output: </strong>44</span>
+</pre>
+
+<p>&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Your Task:</strong><br>
+You don't need to read or print anything. Your task is to complete the function&nbsp;<strong>binary_to_decimal()</strong>&nbsp;which takes the binary number as string input parameter and returns its decimal equivalent.</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Expected Time Complexity:&nbsp;</strong>O(K * Log(K)) where K is number of bits&nbsp;in binary number.<br>
+<strong>Expected Space Complexity:&nbsp;</strong>O(1)</span><br>
+&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+1 &lt;= number of bits in binary number&nbsp;&nbsp;&lt;= 16</span></p>
+ <p></p>
+            </div>

C++/Geeks4Geeks/Binary number to decimal number/binary-number-to-decimal-number.cpp

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+// { Driver Code Starts
+#include<bits/stdc++.h>
+using namespace std;
+
+ // } Driver Code Ends
+class Solution
+{
+	public:
+		int binary_to_decimal(string str)
+		{
+		    // Code here.
+		    int decimal=0;
+		    int power = 1;
+		    for(int i=str.length()-1;i>=0;i--){
+		        int a = str[i]-'0';
+		        decimal += a*power;
+		        power *= 2;
+		    }
+		    return decimal;
+		}
+};
+
+// { Driver Code Starts.
+int main(){
+    int T;
+    cin >> T;
+    while(T--)
+    {
+    	string str;
+    	cin >> str;
+    	Solution ob;
+    	int  ans = ob.binary_to_decimal(str);
+    	cout << ans <<"\n";
+    }
+	return 0;
+}  // } Driver Code Ends

C++/Geeks4Geeks/Bitonic Point/README.md

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+# Bitonic Point
+## Easy 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given an array <strong>arr</strong> of <strong>n</strong> elements which is first increasing and then may be decreasing,&nbsp;find the maximum element in the array.<br>
+<strong>Note: </strong>If the array is increasing then just print then <strong>last element</strong> will be the maximum value.</span></p>
+
+<p><span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong> 
+n = 9
+arr[] = {1,15,25,45,42,21,17,12,11}
+<strong>Output:</strong> 45
+<strong>Explanation:</strong> Maximum element is 45.</span></pre>
+
+<p><span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong> 
+n = 5
+arr[] = {1, 45, 47, 50, 5}
+<strong>Output:</strong> 50
+<strong>Explanation:</strong> Maximum element is 50.</span></pre>
+
+<p><span style="font-size:18px"><strong>Your Task:&nbsp;&nbsp;</strong><br>
+You don't need to read input or print anything. Your task is to complete the function&nbsp;<strong>findMaximum()</strong>&nbsp;which takes the array&nbsp;<strong>arr[], </strong>and<strong> n</strong><strong>&nbsp;</strong>as parameters and returns an integer&nbsp;denoting&nbsp;the answer.<br>
+<br>
+<strong>Expected Time Complexity:</strong>&nbsp;O(logn)<br>
+<strong>Expected Auxiliary Space:</strong>&nbsp;O(1)</span></p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+3 ≤ n ≤ 10<sup>6</sup><br>
+1 ≤ arr<sub>i</sub> ≤ 10<sup>6</sup></span></p>
+
+<p>&nbsp;</p>
+ <p></p>
+            </div>

C++/Geeks4Geeks/Bubble Sort/README.md

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+# Bubble Sort
+## Easy 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given an Integer <strong>N</strong> and a list <strong>arr</strong>. Sort the array using bubble sort algorithm.</span><br>
+<span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input</strong>: 
+N = 5
+arr[] = {4, 1, 3, 9, 7}
+<strong>Output</strong>: 
+1 3 4 7 9</span>
+</pre>
+
+<p><span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input</strong>:
+N = 10 
+arr[] = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}
+<strong>Output</strong>: 
+1 2 3 4 5 6 7 8 9 10</span>
+</pre>
+
+<div><br>
+<strong><span style="font-size:18px">Your Task:&nbsp;</span></strong></div>
+
+<div><span style="font-size:18px">You don't have to read input or print anything. Your task is to complete the function <strong>bubblesort()</strong> which takes the array and it's size as input and sorts the array using bubble sort algorithm.</span></div>
+
+<div><br>
+<span style="font-size:18px"><strong>Expected Time Complexity:</strong>&nbsp;O(N^2).<br>
+<strong>Expected Auxiliary Space:</strong>&nbsp;O(1).</span></div>
+
+<p><br>
+<span style="font-size:18px"><strong>Constraints:</strong><br>
+1 &lt;= N &lt;= 10<sup>3</sup><br>
+1 &lt;= arr[i] &lt;= 10<sup>3</sup></span></p>
+ <p></p>
+            </div>

C++/Geeks4Geeks/Bubble Sort/bubble-sort.cpp

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+// { Driver Code Starts
+//Initial Template for C++
+
+// C program for implementation of Bubble sort
+#include <stdio.h>
+
+// swapping the elements
+void swap(int *xp, int *yp)
+{
+    int temp = *xp;
+    *xp = *yp;
+    *yp = temp;
+}
+
+
+ // } Driver Code Ends
+//User function Template for C++
+
+class Solution
+{
+    public:
+    //Function to sort the array using bubble sort algorithm.
+    void bubbleSort(int arr[], int n)
+    {
+        // Your code here  
+        for(int i=0;i<n;i++){
+            for(int j=i+1;j<n;j++){
+                if(arr[i]>arr[j]){
+                    int a = arr[j];
+                    arr[j] = arr[i];
+                    arr[i] = a;
+                }
+            }
+        }
+    }
+};
+
+
+// { Driver Code Starts.
+
+/* Function to print an array */
+void printArray(int arr[], int size)
+{
+    int i;
+    for (i=0; i < size; i++)
+        printf("%d ", arr[i]);
+    printf("\n");
+}
+
+// Driver program to test above functions
+int main()
+{
+    int arr[1000],n,T,i;
+
+    scanf("%d",&T);
+
+    while(T--){
+
+    scanf("%d",&n);
+
+    for(i=0;i<n;i++)
+      scanf("%d",&arr[i]);
+
+    Solution ob;  
+
+    ob.bubbleSort(arr, n);
+    printArray(arr, n);
+    }
+    return 0;;
+}  // } Driver Code Ends

C++/Geeks4Geeks/C++ Array (print an element) _ Set 2/README.md

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+# C++ Array (print an element) | Set 2
+## School 
+<div class="problem-statement">
+                <p></p><p><span style="font-size:18px">Given an array <strong>A[]</strong> of <strong>N</strong> integers and an index <strong>Key</strong>. Your task is to print the element present at index key in the array.</span></p>
+
+<p>&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Example 1:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong></span><span style="font-size:18px">
+5 2</span>
+<span style="font-size:18px">10 20 30 40 50
+<strong>Output:</strong>
+30</span></pre>
+
+<p>&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Example 2:</strong></span></p>
+
+<pre><span style="font-size:18px"><strong>Input:</strong></span>
+<span style="font-size:18px">7 4</span>
+<span style="font-size:18px">10 20 30 40 50 60 70</span>
+<span style="font-size:18px"><strong>Output:</strong></span>
+<span style="font-size:18px">50</span></pre>
+
+<p>&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Your Task:&nbsp;&nbsp;</strong><br>
+You don't need to read input or print anything. Your task is to complete the function&nbsp;<strong>findElementAtIndex()</strong>&nbsp;which takes the array <strong>A[]</strong>, its size <strong>N </strong>and an integer <strong>Key </strong>as inputs and returns the element present at index Key.</span></p>
+
+<p><br>
+<span style="font-size:18px"><strong>Expected Time Complexity:</strong> O(1)<br>
+<strong>Expected Auxiliary Space:</strong> O(1)</span></p>
+
+<p>&nbsp;</p>
+
+<p><span style="font-size:18px"><strong>Constraints:</strong><br>
+1 ≤ N ≤ 100<br>
+0 ≤ Key ≤ N - 1</span><br>
+<span style="font-size:18px">1 ≤ A[i] ≤ 100</span></p>
+ <p></p>
+            </div>