Dibs commit from Auz

Ashu Yadav/Perm of a string.cpp

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+//This program will solve the mentioned issue number #72
+//Credits:Ashu Yadav
+//Finding all permutation that can be formed by a user entered string.
+//Im considering small input string for the current moment of time
+
+#include<bits/stdc++.h>
+using namespace std;
+
+void ans(string s,int beg,int end){
+    //if the beg and ending index of the string are same in our passed rec fun then print the string
+    //this is our base case that will stop the rec call
+    if(beg==end){
+        cout<<s<<endl;
+    }else{
+        //traversing the string from beg index to last
+        for(int i=beg;i<=end;i++){
+            //using ibuilt swap function,you can create your own swap function also using a temp variable
+            swap(s[beg],s[i]);
+            //call to our fun
+            ans(s,beg+1,end);
+            swap(s[beg],s[i]);
+        }
+    }
+}
+
+int main(){
+    //user entered string is stored
+    string s;
+    cin>>s;
+    //length of the string 
+    int len=s.length();
+    //our recursive function called
+    ans(s,0,len-1);
+    return 0;
+}

Ashu Yadav/count of pairs.cpp

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+// We have to count the number of pairs whose sum is divisible by the number pased by the user
+//iterative search can give TLE so we will be using another approach here
+//Credits:Ashu Yadav
+
+#include<bits/stdc++.h>
+using namespace std;
+//logic
+int ans(int a[],int n,int k){
+    //array to store remainder
+    int occ[k]={0};
+    //remainder 
+    for(int i=0;i<n;i++)occ[a[i]%k]++;
+    //count initiated
+    int count=occ[0]*(occ[n-1])/2;
+    //counting all apirs using n*n-1/2 as number of pairs
+    for(int i=1;i<=k/2 && i!=(k-i);i++){
+        count+=occ[i]*occ[k-i];
+        if(k%2==0)count+=(occ[k/2]*occ[k/2-1])/2;
+    }
+    return count;
+}   
+
+int main(){
+    //size of array
+    int n;
+    cin>>n;
+    int a[n];
+    for(int i=0;i<n;i++)cin>>a[i];
+    int k;
+    cin>>k;
+    //ans
+    cout<<ans(a,n,k)<<endl;
+    return 0;
+}

Ashu Yadav/max height of tree.cpp

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+//we have to find the max height of the binary tree 
+//We will use the recursive aproach to find the height of the tree
+//Credits:Ashu Yadav
+
+#include<bits/stdc++.h>
+using namespace std;
+//Finding the height of the tree
+int height(node* node)  
+{  
+    //base case to terminate the rec call
+    if (node == NULL)  
+        return 0;  
+    else
+    {  
+        int lefth = height(node->left);  
+        int righth = height(node->right);  
+
+        if (lefth > righth)  
+            return(lefth + 1);  
+        else return(righth + 1);  
+    }  
+}  
+//A node class to define the char of a node
+class node  
+{  
+    public: 
+    int data;  
+    //pointers for pointing the left subtree and rightsubtree nodes
+    node* left;  
+    node* right;  
+};
+
+//for making new nodes
+node* nnode(int value)  
+{  
+    node* Node = new node(); 
+    Node->data = value;  
+    Node->left = NULL;  
+    Node->right = NULL;  
+
+    return(Node);  
+}  
+int main(){
+    //Here we can take the data from the user of can pre define the data 
+    //use "nnode" function to make new node and assigning values to it
+
+    //finally we can call height function defined to get the height of the tree
+    cout<<height(root)<<endl;
+    return 0;
+}

Ashu Yadav/traversal in tree.cpp

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+//Implementation of various tree traversal techniques
+//Credtis:Ashu Yadav
+
+#include<bits/stdc++.h>
+using namespace std;
+//Char of node
+struct Node 
+{ 
+    int data; 
+    struct Node* left, *right; 
+    Node(int data) 
+    { 
+        this->data = data; 
+        left = right = NULL; 
+    } 
+}; 
+//function to print postorder
+void Postorder(struct Node *node){
+    //base case to terminate rec
+    if(node==NULL)return;
+    //left subtree
+    Postorder(node->left);
+    //right subtree
+    Postorder(node->right);
+    //print data of node
+    cout<<node->data<<" "; 
+}
+
+//function to print preorder
+void Preorder(struct Node *node){
+    //base case to terminate rec
+    if(node==NULL)return;
+    //print data of node
+    cout<<node->data<<" ";
+    //left subtree
+    Preorder(node->left);
+    //right subtree
+    Preorder(node->right); 
+}
+
+//function to print inorder
+void Inorder(struct Node *node){
+    //base case to terminate rec
+    if(node==NULL)return;
+    //left subtree
+    Inorder(node->left);
+    //print data of node
+    cout<<node->data<<" ";
+    //right subtree
+    Inorder(node->right); 
+}
+int main(){
+    //Here we can take the data from the user of can pre define the data 
+    //use "nnode" function to make new node and assigning values to it
+
+    //prints preorder traversal
+    cout<<Preorder(root)<<endl;
+    //prints postorder traversal
+    cout<<Postorder(root)<<endl;
+    //prints inorder traversal of the tree
+    cout<<Inorder(root)<<endl;
+    return 0;
+}