void arrange(int *arr, int n)
{
int i = 0;
while(i < n)
{
// Place the next element on the left side.
arr[i >>1] = ++i;
if (i >= n) break;
// Place the next element on the right side.
arr[n - (i >>1) - 1] = ++i;
}
}
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Arrange Numbers in the Array
Easy
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16 upvotes
Problem statement
You have been given an empty array(ARR) and its size N. The only input taken from the user will be N and you need not worry about the array.
Your task is to populate the array using the integer values in the range 1 to N(both inclusive) in the order - 1,3,5,.......,6,4,2.
You need not print the array. You only need to populate it.
Detailed explanation ( Input/output format, Notes, Images )
Input Format :
The first line contains an Integer 't' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first and the only line of each test case or query contains an integer 'N'.
For each test case, print the elements of the array/list separated by a single space.
Output for every test case will be printed in a separate line.
Constraints :
1 <= t <= 10^2
0 <= N <= 10^4
Time Limit: 1sec
Sample Input 1 :
1
6
Sample Output 1 :
1 3 5 6 4 2
Explanation of Sample Input 1 :
Since the value of N is 6, the number will be stored in the array in such a fashion that 1 will appear at 0th index, then 2 at the last index, in a similar fashion 3 is stored at index 1. Hence the array becomes 1 3 5 6 4 2.
Sample Input 2 :
2
9
3
Sample Output 2 :
1 3 5 7 9 8 6 4 2
1 3 2
Arrange Numbers in the Array
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Interview problems
Using only 1 loop
#include<bits/stdc++.h>
void arrange(int *arr, int n)
{
//Write your code here
int x = 1 , y = n , mid = ceil(n/2) -1;
if (n % 2 == 1) {
y = n - 1;
mid++;
}
bool flag = 1;
for (int i = 0; i<n; i++)
{
arr[i] = x;
if (i<mid)
x+=2;
else if (flag)
{
x = y;
flag = 0;
}
else
x -= 2;
}
}
6 views
0 replies
0 upvotes
Interview problems
c++
void arrange(int *arr, int n)
{
//Write your code here
//Write your code here
if(n%2==0){
for(int i=0;i<n;i++){
if(i<n/2)
arr[i]=2*i+1;
else
arr[i]=(n-i)*2;
}
}
else{
for(int i=0;i<n;i++){
if(i<=n/2)
arr[i]=2*i+1;
else
arr[i]=(n-i)*2;
}
}
}
120 views
0 replies
1 upvote
Interview problems
Short and easy solution
int count=1;
for(int i=0; i<n/2; i++){
arr[i]=count++;
arr[n-1-i]=count++;
}
if(n%2!=0){
arr[n/2]=count;
}
124 views
0 replies
1 upvote
Interview problems
arrange numbers in array
public class Solution {
public static void arrange(int[] arr, int n) {
//Your code goes here
int i=0;
int j=n-1;
int k=1;
while(i<j){
arr[i]=k;
k++;
arr[j]=k;
k++;
i++;
j--;
}
if(i==j) {
arr[i]=k;
}
}
}
99 views
0 replies
0 upvotes
easy jaava solution
public class Solution {
public static void arrange(int[] arr, int n) {
//Your code goes here
int a=1,i=0;
while(a<=n){
arr[i]=a;
a+=2;
i++;
}
if(n%2==0){
int b=n;
while((i<n)&&(b>0)){
arr[i]=b;
i++;
b-=2;
}
}
else{
int b=n-1;
while((i<n)&&(b>0)){
arr[i]=b;
b-=2;
i++;
}
}
}
}
74 views
0 replies
1 upvote
Interview problems
Solution in python
def arrange(arr, n) :
if n <= 2:
return
val=0
start = 0
end = n-1
while start<=end:
if val%2==0:
arr[start]=val+1
val += 1
start += 1
else:
arr[end]=val+1
val += 1
end -= 1
#Your code goes here
106 views
0 replies
0 upvotes
Interview problems
C++ Solution >>
void arrange(int *arr, int n)
{
int val=1;
int start = 0 ,end = n-1;
while(start<=end){
if(val%2==1){
arr[start]=val;
val++;
start++;
}
else{
arr[end]=val;
val++;
end--;
}
}
} 202 views
0 replies
0 upvotes
1
2
C++ (g++ 5.4)
Console