Dictionary Order(Larger) Solution Added

Author: akashmodak97

Hacker Blocks/Dictionary Order(Larger).cpp

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+/* Hacker Blocks */
+/* Title - Dictionary Order(Larger) */
+/* Created By - Akash Modak */
+/* Date - 14/07/2020 */
+
+// Take as input str, a string. Write a recursive function which prints all the words possible by rearranging the characters of this string which are in dictionary order larger than the given string.
+// The output strings must be lexicographically sorted.
+
+// Input Format
+// Single line input containing a string
+
+// Constraints
+// Length of string <= 10
+
+// Output Format
+// Display all the words which are in dictionary order larger than the string entered in a new line each. The output strings must be sorted.
+
+// Sample Input
+// cab
+
+// Sample Output
+// cba
+// Explanation
+// The possible permutations of string "cab" are "abc" , "acb" ,"bac" , "bca" , "cab" and "cba" . Only one of them is lexicographically greater than "cab". We only print "cba".
+
+#include <iostream>
+#include<string.h>
+#include<set>
+#include<iterator>
+
+using namespace std;
+
+ char temp[1000];
+
+ set<string>s;
+
+void permute(string in, int i)
+{
+    if(in[i]=='\0')
+    {
+       // cout<<in<<endl;
+
+         s.insert(in); //THIS IS NOT WORKING FINE
+
+        return;
+    }
+
+    for(int j=i;in[j]!='\0';j++)
+    {
+       swap(in[i],in[j]);
+       permute(in,i+1);
+       swap(in[i],in[j]);
+    }
+
+
+}
+
+
+int main()
+{
+   string in;
+   cin>>in;
+
+   string temp;
+
+   temp=in;
+
+
+   permute(in,0);
+
+   set<string> :: iterator itr;
+
+
+   for(itr=s.begin();itr!=s.end();itr++)
+   {
+     if(*itr>temp)
+     {
+         cout<<*itr<<endl;
+     }
+   }
+
+    return 0;
+}