String class member functions

-

string a;

a.size() or a.length();              //       to find the length of string.

reverse(a.begin(),a.end());    //        to reverse the string

a.swap(b)         //        to swap string a with b.



To enter a multi word string like"my name is ABC";

string s;
getline(cin,s);
and it can simply be printed on screen using cout.

To enter a multiple line string;

#include <bits/stdc++.h>
using namespace std;
int main() {
    int t;
    string s,a;
    cin>>t;
    while(t--)
{
 cin>>a ;
 s+=a+"\n";     // this (\n) make a multi line string.
}
   cout<<s;
}

INPUT 1:
5
XO|XX
XX|XX
XO|OX
XO|OO
OX|XO

OUTPUT 1:
XO|XX
XX|XX
XO|OX
XO|OO
OX|XO

INPUT 2:
3
RAM Rajesh Ramesh

OUTPUT 2:
RAM
Rajesh 
Ramesh

To find the position of character and replace it with other;

string st_name;
st_name.find("character/ multiple adjacent characters");
//this predefined function returns the first occurred position of the searched element.
Note:The index of the first character is 0 (not 1).
st_name.replace(position, len, str);
// position -- It is the starting insertion point,
  len= the length upto which the insertion is to be done,
  str= string object or value that is is to be placed;



#include<bits/stdc++.h>
using namespace std;
int main() 
{ 
    string s = "The world is so beautiful"; 
string a = "hi, I am your friend";
    int pos =s.find("so");
    s.replace(pos  , 2 , "gm" );
    cout<<a.find('h')<<" "<<a.find("hi")<<" "<<a.find('i')<<" "<<a.find(" ")<<endl;
    cout<<s<<endl<<pos;
   //position of searched element is also displayed for better understanding.
    return 0; 
} 

OUTPUT :
0 0 1 3
The world is gm beautiful
13

NOTE    positions are allocated starting from (0) and for a multi word string the new line character 
i,e;  \n also occupies one position.
EX-
#include <bits/stdc++.h>
using namespace std;
int main() {
    int t,pos;
    string s,a;
    cin>>t;
    while(t--) cin>>a , s+=a+"\n"; // contructs a multiple line string.
       cout<<s.find("XO")<<endl<<s.find("OO");
} 

INPUT:
5
XO|XX
XX|XX
XO|OX
XO|OO
OX|XO

OUTPUT:
0
21



To convert a character of string to integer.
#include<bits/stdc++.h>
#include<sstream>
using namespace std;
int main()
{
int a,b=2;
string s; char c;
cin>>s;                  
c=s[3];            //Lets see if we can convert the character at index 3 to integer,
a=s[3]-48;     // 48 is subtracted because integers lag characters in ASCII code by 48;
Or
a=s[3]-'0';
Or
istringstream(c)>>a; //this predefined function converts string or character into int.
cout<<a+b;      //we added another int. to (a) to make sure that (a) is converted to int., because int can only be added to int.
}

INPUT:
123454

OUTPUT:
6










Comments

Post a Comment

Popular posts from this blog

First_Come_First_Serve CPU Scheduling

-
#include<stdio.h> #include<conio.h> struct proc {     int id;     int burst;     int at; }; int main() {     int n,i,j,total_waiting_time,waiting_time;     printf("Enter the number of processess: ");     scanf("%d",&n);     struct proc p[n],temp;     f or(i=0;i<n;i++)     {         p[i].id=i;         printf("Enter the arrival time of the process: ");         scanf("%d",&p[i].at);         printf("Enter the burst time of the process: ");         scanf("%d",&p[i].burst);     }     for(i=0;i<n;i++)     {         for(j=i+1;j<n;j++)         {             if(p[j].at<p[i].at)             {                 ...
Read more

Reversing stack Method 2 !! (One Helper Stack only)

-
#include   <bits/stdc++.h> using   namespace   std ; stack < int >  rev ( stack < int >  s ) {      int   size  =  s . size (),  x ,  y ,  t ;      stack < int >  s2 ;      size  -=  1 ;      while  ( size --)     {          t  =  size  +  1 ;          x  =  s . top ();          s . pop ();          while  ( t --)         {              s2 . push ( s . top ());              s . pop ();         }  ...
Read more

Populating Next Right Pointers in Each Node in O(1) space (without queue and level order)

-
Image
Example 1: Input: root = [1,2,3,4,5,6,7] Output: [1,#,2,3,#,4,5,6,7,#] Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level. Example 2: Input: root = [] Output: [] class Solution { public:     Node * connect ( Node * root )     {         Node * p1 = root , * p2 ;         while ( p1 != NULL )         {             p2 = p1 ;             while ( p2 != NULL )             {                 if ( p2 -> left )                     p2 -> left -> next = p2 -> right ;       ...
Read more

Calculate factorial of large numbers !! (Using Arrays)

-
#include   <bits/stdc++.h> using   namespace   std ; #define   max   10000 int   main () {      int   n ,  c ,  j ,  k ,  x ,  i ;      int   a [ max ];      cout   <<   "Enter the number: " ;      cin   >>   n ;      j  =  0 ;  // j will be pointing to new index created to stored carry      a [ 0 ] =  1 ;      for  ( i  =  2 ;  i  <=  n ;  i ++)     {          c  =  0 ;  //carry when each number of array is multiplied by i          //(i is a...
Read more

Multiplication of large numbers (Given in string format)

-
  class   Solution { public:      string   multiply ( string   num1 ,  string   num2 )     {          if  ( num1  ==  "0"  ||  num2  ==  "0" )              return   "0" ;          if  ( num1  ==  "1" )              return   num2 ;          if  ( num2  ==  "1" )              return   num1 ;          if  ( num1 . size () ==  1  &&  num2 . size () ==  1 )         {           ...
Read more

Left View of Binary Tree (Method 1 using recursion)

-
#include   <iostream> #include   <vector> using   namespace   std ; struct   TreeNode {      int   val ;      TreeNode  * left ;      TreeNode  * right ;      TreeNode () :  val ( 0 ),  left ( nullptr ),  right ( nullptr ) {}      TreeNode ( int   x ) :  val ( x ),  left ( nullptr ),  right ( nullptr ) {}      TreeNode ( int   x ,  TreeNode   * left ,  TreeNode   * right ) :                              val ( x ),  left ( left ),  right ( right ) {} }; TreeNode   * BuildTree () {      int   d ;      cin   >>   d ;      if  ( d  == -...
Read more

Check Bracket Sequence

-
#include<bits/stdc++.h> using namespace std; int main() {     int i,f,top=-1,size;     string s;     cin>>s;     size=s.size();     char a[size];     f=1;     for(i=0;i<s.size() && f==1;i++)     {         if(s[i]=='(')         a[++top]='(';         else if(s[i]==')')         {             if(top==-1)             f=0;             else             top-=1;         }     }     if(top!=-1 || f==0)     cout<<"invalid\n";     else     cout<<"Valid"; }
Read more

Image Multiplication

-
https://practice.geeksforgeeks.org/problems/image-multiplication0627/1 Input: 1 / \ 3 2 / \ / \ 7 6 5 4 / \ \ / \ \ 11 10 15 9 8 12 Output: 332 Explanation: Sum = (1*1) + (3*2) + (7*4) + (6*5) + (11*12) + (15*9) = 332 class Solution {     public:     void treeHelper ( Node * a , Node * b , long long int & ans )     {         if (a== NULL ||b== NULL )         return ;         long long int temp=( a -> data * b -> data )%( 1000000007 );         ans=(ans+temp)%( 1000000007 );         treeHelper ( a -> left , b -> right , ans);         treeHelper ( a -> right , b -> left , ans);     }   ...
Read more

Boundary Traversal of binary tree

-
struct Node {     int data;     Node * left, * right; }; class Solution { public:     void call_left ( vector < int > & v , Node * root )     {         if ( root == NULL )             return ;         if ( root ->left == NULL && root ->right == NULL )             return ; // don't push the leaf nodes.         v . push_back ( root ->data);         if ( root ->left)             call_left ( v , root ->left);         else             call_left ( v , root ->right);     }     void call_leaves ( vector < int > & v , Node * root )     {         if ( root == NULL )             return ;   ...
Read more

BST to greater sum tree

-
  // Input: //            2 //          /    \ //         1      6 //               /  \ //              3    7 // Output: 18 16 13 7 0 (InOrder) // Explanation: // Every node is replaced with the sum of nodes greater than itself. // The resultant tree is: //                16 //              /    \ //            18       7 //                   /   \ //                  13    0 //Bhaskar Varshney #include <iostream> #include <queue> using namespace std; struct TreeNode {     int val;     TreeNode *left;     TreeNode *rig...
Read more