You have been given an integer array/list(ARR) of size 'N'. It only contains 0s, 1s and 2s. Write a solution to sort this array/list.
Note :
Try to solve the problem in 'Single Scan'. ' Single Scan' refers to iterating over the array/list just once or to put it in other words, you will be visiting each element in the array/list just once.
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 line of each test case contains an Integer 'N' denoting the size of the array/list.
The second line of each test case contains 'N' space-separated Integers denoting the array/list.
For each test case/query, print the sorted array/list(ARR) as space-separated Integers.
Output for every test case will be printed in a separate line.
You need to change in the given array/list itself. Hence, no need to return or print anything.
1 <= T <= 10
1 <= N <= (5 * (10 ^ 5))
0 <= ARR[i] <= 2
Where 'N' is the size of the given array/list.
And, ARR[i] denotes the i-th element in the array/list.
Time Limit: 1sec
2
6
0 1 2 2 1 0
7
0 1 2 1 2 1 2
0 0 1 1 2 2
0 1 1 1 2 2 2
2
7
2 2 2 1 1 1 0
6
2 1 2 0 1 0
0 1 1 1 2 2 2
0 0 1 1 2 2
Can try to find the sorting algorithm of best time Complexity.
Use any good sorting algorithm like Merge Sort, Quick Sort or inbuilt sorting function of different languages.
O(N*log(N)), where ‘N’ is the size of the array.
We are using inbuilt sort algorithm which has Overall Time Complexity O(N*log(N))
O(1), As we are using constant space.
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Interview problems
Dutch national flag algorithm
We can have 3 pointers: low, mid and high. low and high are at the either ends of the array and mid traverses through it. If element at mid == 0, we swap it with low pointer, if element at mid == 1, we simply move forward else it is 2 for which we swap it with high pointer. The loop runs till mid≤high.
Interview problems
why is the program printing an extra 0 at the beginning of the output
#include <bits/stdc++.h>
void swap(int &m,int &n)
{
int temp=m;
m=n;
n=temp;
}
void sort012(int *arr, int n)
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(arr[i]<arr[j])
swap(arr[i],arr[j]);
}
}
for (int k = 0; k < n; k++) {
cout << arr[k] << " ";
break;
}
}
Interview problems
sort012
#include <bits/stdc++.h>
void sort012(int *arr, int n)
{
// Write your code here
int temp1,temp2;
for(int i=0;i<n-1;i++)
{
if(arr[i]>arr[i+1])
{
temp1 = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp1;
}
while(i>0 && arr[i]<arr[i-1])
{
temp2 = arr[i];
arr[i] = arr[i-1];
arr[i-1] = temp2;
i--;
}
}
}
Interview problems
Sort 0 1 2
import java.util.* ;
import java.io.*;
public class Solution
{
public static void sort012(int[] arr)
{
int low=0,mid=0,high=arr.length-1;
while(mid<=high){
if(arr[mid]==0){
int temp=arr[low];
arr[low]=arr[mid];
arr[mid]=temp;
low++;
mid++;
}else if(arr[mid]==1){
mid++;
}else{
int temp2=arr[mid];
arr[mid]=arr[high];
arr[high]=temp2;
high--;
}
}
}
}
Interview problems
My rookie solution
//I created new array and I would traverse around the given array and if I saw 0 then I would put it in consecutively in indexes starting from 0 and if I saw 2 then I would put that backwardly starting from index N-1 and remaining places I would fill them with 1.
#include <bits/stdc++.h>
void sort012(int *arr, int n)
{
int new_arr[n];
int j=0, k=n-1;
for(int i=0; i<n; i++){
if(arr[i]==0){
new_arr[j]=0;
j++;
}
if(arr[i]==2){
new_arr[k]=2;
k--;
}
}
for(int l=j; l<k+1; l++){
new_arr[l]=1;
}
for(int i=0; i<n; i++){
arr[i]=new_arr[i];
}
}
Interview problems
using DNF - Dutch National Flag Algorithm
int low = 0;
int mid = 0;
int high = arr.length-1;
while(mid<=high) {
if (arr[mid] == 0) {
int temp = arr[low];
arr[low] = arr[mid];
arr[mid] = temp;
low++;
mid++;
}
else if (arr[mid] == 1) {
mid++;
}
else if (arr[mid] == 2) {
int temp = arr[high];
arr[high] = arr[mid];
arr[mid] = temp;
high--;
}
}
Interview problems
c++ shortest way to solve
#include <bits/stdc++.h>
#include<algorithm>
void sort012(int *arr, int n)
{
// Write your code here
sort(arr , arr+n);
//this function sort the array from start to end where //n represents no. of elements
}
Interview problems
solve in "single line " Check it out.............
import java.util.* ;
import java.util.concurrent.ArrayBlockingQueue;
import java.io.*;
public class Solution
{
public static void sort012(int[] arr)
{
//Write your code here
Arrays.sort(arr);
}
}
Interview problems
o(n) complexity
int x=0,y=0,z=0,j=0;
for (int i = 0; i < n; i++)
{
if(arr[i]==0) x++;
else if (arr[i]==1) y++;
else z++;
}
while(x){
arr[j++] = 0;
x--;
}
while(y){
arr[j++] = 1;
y--;
}
while(z){
arr[j++] = 2;
z--;
}
Interview problems
sort 0 1 2
#include <bits/stdc++.h>
void sort012(int *arr, int n)
{
int c1=0;
int c2=0;
int c3=0;
for(int i=0;i<n;i++){
if(arr[i]==0){
c1++;
}
else if(arr[i]==1){
c2++;
}
else{
c3++;
}
}
int index=0;
for(int i=0;i<c1;i++){
arr[index]=0;
index++;
}
for(int i=0;i<c2;i++){
arr[index]=1;
index++;
}
for (int i = 0; i < c3; i++) {
arr[index] = 2;
index++;
}
}
void print(int arr[],int n){
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
