gfg

Author: jaygajera17

convert_to_lowercase.cpp

@@ -0,0 +1,58 @@
+
+/*
+Link:- https://practice.geeksforgeeks.org/problems/java-convert-string-to-lowercase2313/1
+The task is to convert characters of string to lowercase.
+
+
+Input: S = "ABCddE"
+Output: "abcdde"
+Explanation: A, B, C and E are converted to
+a, b, c and E thus all uppercase characters
+of the string converted to lowercase letter.
+
+*/
+
+string toLower(string s)
+{
+    // code here
+    string ans = "";
+    for (int i = 0; i < s.length(); i++)
+    {
+        if (s[i] >= 'A' && s[i] <= 'Z')
+            s[i] = s[i] + 32;
+        ans = ans + s[i];
+        // ans=ans+ (char)tolower(s[i]);
+    }
+    return ans;
+}
+
+/*
+
+   //Using tolower() function
+
+    string toLower(string s) {
+
+        string ans="";
+        for(int i=0;i<s.length();i++)
+        {
+          ans=ans+ (char)tolower(s[i]);
+        }
+        return ans;
+    }
+
+*/
+
+/*
+    /***** bonus *****
+
+  //Convert to CamelCase
+ s[0]=s[0]-32;
+    for(int i=0;i<s.size();i++)
+    {
+      if(s[i]==' ')
+        {
+          s[i+1]=s[i+1]-32;
+        }
+    }
+    return s;
+ */

display_longest_name.cpp

@@ -0,0 +1,30 @@
+/* 
+Given a list of names, display the longest name.
+
+link:- https://practice.geeksforgeeks.org/problems/display-longest-name0853/1
+
+Input:
+N = 5
+names[] = { "Geek", "Geeks", "Geeksfor",
+  "GeeksforGeek", "GeeksforGeeks" }
+
+Output:
+GeeksforGeeks
+
+
+*/
+
+    int l=0;
+    string ans="";
+    string longest(string names[], int n)
+    {
+        for(int i=0;i<n;i++)
+        {
+            if(names[i].length()>=l)
+            {
+                l=names[i].length();
+                ans=names[i];
+            }
+        }
+        return ans;
+    }

equilibrium_point.cpp

@@ -0,0 +1,63 @@
+/* 
+    NOTE:- it is imp ARRAY QUESTION NOT STRING 
+    link:- https://practice.geeksforgeeks.org/problems/equilibrium-point-1587115620/1
+
+    problem:- Given an array A of n positive numbers. The task is to find the first Equilibium Point in the array. 
+Equilibrium Point in an array is a position such that the sum of elements before it is equal to the sum of elements after it.
+
+Input: 
+n = 5 
+A[] = {1,3,5,2,2} 
+Output: 3 
+Explanation: For second test case 
+equilibrium point is at position 3 
+as elements before it (1+3) = 
+elements after it (2+2). 
+
+Input:
+n = 1
+A[] = {1}
+Output: 1
+Explanation:
+Since its the only element hence
+its the only equilibrium point.
+
+*/
+
+// { Driver Code Starts
+#include <iostream>
+using namespace std;
+
+
+ // } Driver Code Ends
+class Solution{
+    public:
+    // Function to find equilibrium point in the array.
+    // a: input array
+    // n: size of array
+    int equilibriumPoint(long long arr[], int n) {
+  int t_sum=0,sum1=0,sum2=0;
+  if(n==1)
+{
+       return 1;
+}
+  for(int i=0;i<n;i++)
+{
+       t_sum+=arr[i];
+}
+  for(int i=1;i<n;i++)
+{
+       sum1=sum1+arr[i-1];
+       sum2=t_sum-sum1-arr[i];
+   
+      if(sum1==sum2)
+      {
+          return i+1;
+      }
+     
+}
+  return -1;
+}
+
+};
+

reverse_string.cpp

@@ -16,6 +16,14 @@ string reverseWord(string str){
     {
         s=s+str[i];
     }
+     
+      
     return s;
   //Your code here
 }
+
+              /* using swap */
+//  for(int i=0;i<s.length()/2;i++)
+//      {
+//         swap(s[i],s[s.length()-i-1]);
+//      }

reversing_the_vowel.cpp

@@ -0,0 +1,44 @@
+/*
+Link:-https://practice.geeksforgeeks.org/problems/reversing-the-vowels5304/1
+problem:- Given a string consisting of lowercase english alphabets, reverse only the vowels present in it and print the resulting string.
+
+Input:
+S = "geeksforgeeks"
+Output: geeksforgeeks
+Explanation: The vowels are: e, e, o, e, e
+Reverse of these is also e, e, o, e, e.
+
+Input: 
+S = "practice"
+Output: prectica
+Explanation: The vowels are a, i, e
+Reverse of these is e, i, a.
+
+*/
+
+ string modify (string s)
+        {
+            //code here.
+            string vowel="";
+            for(int i=0;i<s.length();i++)
+            {
+                if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u')
+                {
+                    vowel+=s[i];
+                }
+            }
+            
+            
+            int n=vowel.length();
+            
+        for(int i=0;i<s.length();i++)
+            {
+                if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u')
+                {
+                   s[i]=vowel[n-1];
+                   n--;
+                }
+            }
+            
+            return s;
+        }