• Skip to primary navigation
  • Skip to content
  • Skip to primary sidebar
  • Skip to secondary sidebar

GoHired

Interview Questions asked in Google, Microsoft, Amazon

Join WeekEnd Online Batch from 4-April-2020 on How to Crack Coding Interview in Just 10 Weeks : Fees just 20,000 INR

  • Home
  • Best Java Books
  • Algorithm
  • Internship
  • Certificates
  • About Us
  • Contact Us
  • Privacy Policy
  • Array
  • Stack
  • Queue
  • LinkedList
  • DP
  • Strings
  • Tree
  • Mathametical
  • Puzzles
  • Graph

Printing intermediate Integers between one element & next element of array

February 16, 2018 by Dhaval Dave

Problem : Printing intermediate Integers

User entered numbers are set into array.Intermediate integer of each pair of successive elements to be printed.

Input :
Input of 1st line is number of element in array,inputs of 2nd line are array elements.
Output :
Output of each line is intermediate integers between each two succesive array element.
Example :

Input Output
3
2 6 8 3
3 4 5
7
4 5 6 7
3
8 8 9
"no  intermediate element in between these 8 & 9 elements"
"no  intermediate element in between these 9 & 9 elements"
 Solution : 

Logic for Printing intermediate Integers 

1. Entered numbers kept stored in array                             
2. For loop executes for each pair of successive array elements.
i) If two successive array elements are same or difference between two successive array elements is one ,then no intermediate integer presence message is printed . 
ii) Lesser number is identified in between two numbers of each pair.
iii) printing starts from the number one more than lesser number ,this number incremented by one,again printed,this continues upto the number one less than grater number between pair.

Code : 

#include"iostream";
using namespace std;
main()
{
 int n;
 cout<<"Enter number of Integer in array "<<endl;
 cin>>n;
 int num[n];
 cout<<"Enter elements of array "<<endl;
 for(int p=0;p<n;p++)
 cin>>num[p];
 for(int k=0;k<n-1;k++)
 {
   if((num[k]-num[k+1]==1) || (num[k+1]-num[k]==1) || (num[k]==num[k+1]))
   {
   cout<<"No intermediate element in between these "<<num[k]<<" & "<<num[k+1]<<" elements "<<endl;
   }
   else if(num[k]<num[k+1])
   {
     cout<<"Element between "<<num[k]<<" and "<<num[k+1]<<" is/are ";
     for(int a=num[k];a<num[k+1]-1;)
     {
        cout<<++a<<" ";   //printing intermediate element 
     }
     cout<<endl;
   }
   else if(num[k]>num[k+1])
   {
     cout<<"Element between "<<num[k]<<" and "<<num[k+1]<<" is/are ";
     for(int a=num[k+1];a<num[k]-1;)
       {
         cout<<++a<<" ";   //printing intermediate element
       }
       cout<<endl;
    }
  }
}

Click here to see running code on Ideone with input 4 2 6 8 3

Click here to see running code on Ideone with input 3 8 9 9

Explanation : 

Entered 1st input set’s numbers are stored in array ‘num’ like below –


First (2,6) ,then (6,8),then (8,3) pair get considered.Intermediate values are printed.
Entered 2nd input set’s numbers are stored in array ‘num’ like below –


First (8,9) ,then (9,9) pair get considered.No intermediate element present message is printed.

Another Logic and Code : 

#include"iostream";
using namespace std;
main()
{
int n,a,b,large,small;
cout<<"Enter number of Integer in array "<<endl;
cin>>n;
int num[n];
cout<<"Enter element of array "<<endl;
for(int p=0;p<n;p++)
cin>>num[p];
for(int k=0;k<n-1;k++)
{
 a=num[k];
 b=num[k+1];
 if((num[k]-num[k+1]==1) || (num[k+1]-num[k]==1) || (num[k]==num[k+1]))
 {
 cout<<"No intermediate element present in between these "<<num[k]<<" & "<<num[k+1]<<" elements";
 }
else
{
large=a>b?a:b;   //identifying larger number of pair 
small=a<b?a:b;   //identifying smaller number of pair
cout<<"Element between "<<num[k]<<" and "<<num[k+1]<<" is/are ";
for(int p=small;p<large-1;)
{
 cout<<++p<<" ";   //printing intermediate element
}
}
cout<<endl;
}
}

Click here to see running code on Ideone with input 4 2 6 8 3
Click here to see running code on Ideone with input 3 8 9 9

Explanation : 

If two successive array elements are same or difference between two successive array elements is one ,then no intermediate integer presence message is printed .Larger & smaller number in pair are identified by ternary operator ,then printing starts from the number one more than small ,continues upto the number one less than larger .

Similar Articles

Filed Under: Uncategorized

Reader Interactions

Primary Sidebar

Join WeekEnd Online/Offline Batch from 4-April-2020 on How to Crack Coding Interview in Just 10 Weeks : Fees just 20,000 INR

Join WeekEnd Online/Offline Batch from 4-April-2020

WhatsApp us

Secondary Sidebar

Custom Search

  • How I cracked AMAZON
  • LeetCode
  • Adobe
  • Amazon
  • Facebook
  • Microsoft
  • Hacker Earth
  • CSE Interview

Top Rated Questions

Add Sub Multiply very large number stored as string

N Petrol bunks or City arranged in circle. You have Fuel and distance between petrol bunks. Is it possible to find starting point so that we can travel all Petrol Bunks

Get K Max and Delete K Max in stream of incoming integers

TicTacToe Game As Asked in Flipkart

Serialise Deserialise N-ary Tree

DFS (Depth First Search)

Facebook Interview Question : Interleave List

Fibonacci Hashing & Fastest Hashtable

Max Sum in circularly situated Values

Linked List V/S Binary Search Tree

In Given LinkedList Divide LL in N Sub parts and delete first K nodes of each part

Flipkart SDET Interview Experience

SAP Hiring Off-Campus General Aptitude

Test Cases for Round Function

SAP Off Campus Hiring_ March 2015 Analytical Aptitude

Reversal of LinkedList

ADOBE Aptitude C Language Test

VMWare SDEII Interview

Circular Linked List

Get Minimum element in O(1) from input numbers or Stack

Given a string, find the first character which is non-repetitive

Sort an array according to the order defined by another array

Interfaces in C++ (Abstract Classes in C++)

Binary Tree in Java

Edit Distance ( Dynamic Programming )

Mirror of Tree

Diagonal Traversal of Binary Tree

Common Ancestor in a Binary Tree or Binary Search Tree

simple sql injection

Hackerearth : Counting Subarrays

Copyright © 2025 · Genesis Framework · WordPress · Log in