Sorting using trivial hash function
Last Updated :
19 Apr, 2023
We have read about various sorting algorithms such as heap sort, bubble sort, merge sort and others.
Here we will see how can we sort N elements using a hash array. But this algorithm has a limitation. We can sort only those N elements, where the value of elements is not large (typically not above 10^6).
Examples:
Input : 9 4 3 5 8
Output : 3 4 5 8 9
Explanation of sorting using hash:
- Step 1: Create a hash array of size(max_element), since that is the maximum we will need
- Step 2: Traverse through all the elements and keep a count of number of occurrence of a particular element.
- Step 3: After keeping a count of occurrence of all elements in the hash table, simply iterate from 0 to max_element in the hash array
- Step 4: While iterating in the hash array, if we find the value stored at any hash position is more than 0, which indicated that the element is present at least once in the original list of elements.
- Step 5: Hash[i] has the count of the number of times an element is present in the list, so when its >0, we print those number of times the element.
- If you want to store the elements, use another array to store them in a sorted way.
- If we want to sort it in descending order, we simply traverse from max to 0 and repeat the same procedure.
Below is the implementation of the above approach:
C++
// C++ program to sort an array using hash
// function
#include <bits/stdc++.h>
using namespace std;
void sortUsingHash(int a[], int n)
{
// find the maximum element
int max = *std::max_element(a, a + n);
// create a hash function upto the max size
int hash[max + 1] = { 0 };
// traverse through all the elements and
// keep a count
for (int i = 0; i < n; i++)
hash[a[i]] += 1;
// Traverse upto all elements and check if
// it is present or not. If it is present,
// then print the element the number of times
// it's present. Once we have printed n times,
// that means we have printed n elements
// so break out of the loop
for (int i = 0; i <= max; i++) {
// if present
if (hash[i]) {
// print the element that number of
// times it's present
for (int j = 0; j < hash[i]; j++) {
cout << i << " ";
}
}
}
}
// driver program
int main()
{
int a[] = { 9, 4, 3, 2, 5, 2, 1, 0, 4,
3, 5, 10, 15, 12, 18, 20, 19 };
int n = sizeof(a) / sizeof(a[0]);
sortUsingHash(a, n);
return 0;
}
// Java program to sort an array using hash
// function
import java.util.*;
class GFG {
static void sortUsingHash(int a[], int n)
{
// find the maximum element
int max = Arrays.stream(a).max().getAsInt();
// create a hash function upto the max size
int hash[] = new int[max + 1];
// traverse through all the elements and
// keep a count
for (int i = 0; i < n; i++)
hash[a[i]] += 1;
// Traverse upto all elements and check if
// it is present or not. If it is present,
// then print the element the number of times
// it's present. Once we have printed n times,
// that means we have printed n elements
// so break out of the loop
for (int i = 0; i <= max; i++) {
// if present
if (hash[i] != 0) {
// print the element that number of
// times it's present
for (int j = 0; j < hash[i]; j++) {
System.out.print(i + " ");
}
}
}
}
// Driver code
public static void main(String[] args)
{
int a[] = { 9, 4, 3, 2, 5, 2, 1, 0, 4,
3, 5, 10, 15, 12, 18, 20, 19 };
int n = a.length;
sortUsingHash(a, n);
}
}
// This code contributed by Rajput-Ji
# Python3 program to sort an array
# using hash function
def sortUsingHash(a, n):
# find the maximum element
Max = max(a)
# create a hash function upto
# the max size
Hash = [0] * (Max + 1)
# traverse through all the elements
# and keep a count
for i in range(0, n):
Hash[a[i]] += 1
# Traverse upto all elements and check
# if it is present or not. If it is
# present, then print the element the
# number of times it's present. Once we
# have printed n times, that means we
# have printed n elements so break out
# of the loop
for i in range(0, Max + 1):
# if present
if Hash[i] != 0:
# print the element that number
# of times it's present
for j in range(0, Hash[i]):
print(i, end=" ")
# Driver Code
if __name__ == "__main__":
a = [9, 4, 3, 2, 5, 2, 1, 0, 4,
3, 5, 10, 15, 12, 18, 20, 19]
n = len(a)
sortUsingHash(a, n)
# This code is contributed by Rituraj Jain
// C# program to sort an array using hash
// function
using System;
using System.Linq;
class GFG {
static void sortUsingHash(int[] a, int n)
{
// find the maximum element
int max = a.Max();
// create a hash function upto the max size
int[] hash = new int[max + 1];
// traverse through all the elements and
// keep a count
for (int i = 0; i < n; i++)
hash[a[i]] += 1;
// Traverse upto all elements and check if
// it is present or not. If it is present,
// then print the element the number of times
// it's present. Once we have printed n times,
// that means we have printed n elements
// so break out of the loop
for (int i = 0; i <= max; i++) {
// if present
if (hash[i] != 0) {
// print the element that number of
// times it's present
for (int j = 0; j < hash[i]; j++) {
Console.Write(i + " ");
}
}
}
}
// Driver code
public static void Main(String[] args)
{
int[] a = { 9, 4, 3, 2, 5, 2, 1, 0, 4,
3, 5, 10, 15, 12, 18, 20, 19 };
int n = a.Length;
sortUsingHash(a, n);
}
}
/* This code contributed by PrinciRaj1992 */
<script>
// javascript program to sort an array using hash
// function
function sortUsingHash(a, n) {
// find the maximum element
var max = Math.max.apply(Math, a);
// create a hash function upto the max size
var hash = Array(max + 1).fill(0);
// traverse through all the elements and
// keep a count
for (i = 0; i < n; i++)
hash[a[i]] += 1;
// Traverse upto all elements and check if
// it is present or not. If it is present,
// then print the element the number of times
// it's present. Once we have printed n times,
// that means we have printed n elements
// so break out of the loop
for (i = 0; i <= max; i++) {
// if present
if (hash[i] != 0) {
// print the element that number of
// times it's present
for (j = 0; j < hash[i]; j++) {
document.write(i + " ");
}
}
}
}
// Driver code
var a = [ 9, 4, 3, 2, 5, 2, 1, 0, 4, 3, 5, 10, 15, 12, 18, 20, 19 ];
var n = a.length;
sortUsingHash(a, n);
// This code contributed by Rajput-Ji
</script>
Output0 1 2 2 3 3 4 4 5 5 9 10 12 15 18 19 20
Time Complexity: O(max*n), where max is maximum element and n is the length of given array
Auxiliary Space: O(max)
How to handle negative numbers?
In case the array has negative numbers and positive numbers, we keep two hash arrays to keep a track of positive and negative elements.
Explanation of sorting using hashing if the array has negative and positive numbers:
- Step 1: Create two hash arrays, one for positive and the other for negative
- Step 2: the positive hash array will have a size of max and the negative array will have a size of min
- Step 3: traverse from min to 0 in the negative hash array, and print the elements in the same way we did for positives.
- Step 4: Traverse from 0 to max for positive elements and print them in the same manner as explained above.
Below is the implementation of the above approach:
C++
// C++ program to sort an array using hash
// function with negative values allowed.
#include <bits/stdc++.h>
using namespace std;
void sortUsingHash(int a[], int n)
{
// find the maximum element
int max = *std::max_element(a, a + n);
int min = abs(*std::min_element(a, a + n));
// create a hash function upto the max size
int hashpos[max + 1] = { 0 };
int hashneg[min + 1] = { 0 };
// traverse through all the elements and
// keep a count
for (int i = 0; i < n; i++) {
if (a[i] >= 0)
hashpos[a[i]] += 1;
else
hashneg[abs(a[i])] += 1;
}
// Traverse up to all negative elements and
// check if it is present or not. If it is
// present, then print the element the number
// of times it's present. Once we have printed
// n times, that means we have printed n elements
// so break out of the loop
for (int i = min; i > 0; i--) {
if (hashneg[i]) {
// print the element that number of times
// it's present. Print the negative element
for (int j = 0; j < hashneg[i]; j++) {
cout << (-1) * i << " ";
}
}
}
// Traverse upto all elements and check if it is
// present or not. If it is present, then print
// the element the number of times it's present
// once we have printed n times, that means we
// have printed n elements, so break out of the
// loop
for (int i = 0; i <= max; i++) {
// if present
if (hashpos[i]) {
// print the element that number of times
// it's present
for (int j = 0; j < hashpos[i]; j++) {
cout << i << " ";
}
}
}
}
// driver program to test the above function
int main()
{
int a[] = { -1, -2, -3, -4, -5, -6, 8,
7, 5, 4, 3, 2, 1, 0 };
int n = sizeof(a) / sizeof(a[0]);
sortUsingHash(a, n);
return 0;
}
// Java program to sort an array using hash
// function with negative values allowed.
import java.util.Arrays;
class GFG {
static int absolute(int x)
{
if (x < 0)
return (-1 * x);
return x;
}
static void sortUsingHash(int a[], int n)
{
// find the maximum element
int max = Arrays.stream(a).max().getAsInt();
int min
= absolute(Arrays.stream(a).min().getAsInt());
// create a hash function upto the max size
int hashpos[] = new int[max + 1];
int hashneg[] = new int[min + 1];
// traverse through all the elements and
// keep a count
for (int i = 0; i < n; i++) {
if (a[i] >= 0)
hashpos[a[i]] += 1;
else
hashneg[absolute(a[i])] += 1;
}
// Traverse up to all negative elements and
// check if it is present or not. If it is
// present, then print the element the number
// of times it's present. Once we have printed
// n times, that means we have printed n elements
// so break out of the loop
for (int i = min; i > 0; i--) {
if (hashneg[i] > 0) {
// print the element that number of times
// it's present. Print the negative element
for (int j = 0; j < hashneg[i]; j++) {
System.out.print((-1) * i + " ");
}
}
}
// Traverse upto all elements and check if it is
// present or not. If it is present, then print
// the element the number of times it's present
// once we have printed n times, that means we
// have printed n elements, so break out of the
// loop
for (int i = 0; i <= max; i++) {
// if present
if (hashpos[i] > 0) {
// print the element that number of times
// it's present
for (int j = 0; j < hashpos[i]; j++) {
System.out.print(i + " ");
}
}
}
}
// Driver program to test the above function
public static void main(String[] args)
{
int a[] = { -1, -2, -3, -4, -5, -6, 8,
7, 5, 4, 3, 2, 1, 0 };
int n = a.length;
sortUsingHash(a, n);
}
}
// This code has been contributed by 29AjayKumar
# Python3 program to sort an array using hash
# function with negative values allowed.
def sortUsingHash(a, n):
# find the maximum element
Max = max(a)
Min = abs(min(a))
# create a hash function upto the max size
hashpos = [0] * (Max + 1)
hashneg = [0] * (Min + 1)
# traverse through all the elements and
# keep a count
for i in range(0, n):
if a[i] >= 0:
hashpos[a[i]] += 1
else:
hashneg[abs(a[i])] += 1
# Traverse up to all negative elements
# and check if it is present or not.
# If it is present, then print the
# element the number of times it's present.
# Once we have printed n times, that means
# we have printed n elements so break out
# of the loop
for i in range(Min, 0, -1):
if hashneg[i] != 0:
# print the element that number of times
# it's present. Print the negative element
for j in range(0, hashneg[i]):
print((-1) * i, end=" ")
# Traverse upto all elements and check if
# it is present or not. If it is present,
# then print the element the number of
# times it's present once we have printed
# n times, that means we have printed n
# elements, so break out of the loop
for i in range(0, Max + 1):
# if present
if hashpos[i] != 0:
# print the element that number
# of times it's present
for j in range(0, hashpos[i]):
print(i, end=" ")
# Driver Code
if __name__ == "__main__":
a = [-1, -2, -3, -4, -5, -6,
8, 7, 5, 4, 3, 2, 1, 0]
n = len(a)
sortUsingHash(a, n)
# This code is contributed by Rituraj Jain
// C# program to sort an array using hash
// function with negative values allowed.
using System;
using System.Linq;
class GFG {
static int absolute(int x)
{
if (x < 0)
return (-1 * x);
return x;
}
static void sortUsingHash(int[] a, int n)
{
// find the maximum element
int max = a.Max();
int min = absolute(a.Min());
// create a hash function upto the max size
int[] hashpos = new int[max + 1];
int[] hashneg = new int[min + 1];
// traverse through all the elements and
// keep a count
for (int i = 0; i < n; i++) {
if (a[i] >= 0)
hashpos[a[i]] += 1;
else
hashneg[absolute(a[i])] += 1;
}
// Traverse up to all negative elements and
// check if it is present or not. If it is
// present, then print the element the number
// of times it's present. Once we have printed
// n times, that means we have printed n elements
// so break out of the loop
for (int i = min; i > 0; i--) {
if (hashneg[i] > 0) {
// print the element that number of times
// it's present. Print the negative element
for (int j = 0; j < hashneg[i]; j++) {
Console.Write((-1) * i + " ");
}
}
}
// Traverse upto all elements and check if it is
// present or not. If it is present, then print
// the element the number of times it's present
// once we have printed n times, that means we
// have printed n elements, so break out of the
// loop
for (int i = 0; i <= max; i++) {
// if present
if (hashpos[i] > 0) {
// print the element that number of times
// it's present
for (int j = 0; j < hashpos[i]; j++) {
Console.Write(i + " ");
}
}
}
}
// Driver code
public static void Main(String[] args)
{
int[] a = { -1, -2, -3, -4, -5, -6, 8,
7, 5, 4, 3, 2, 1, 0 };
int n = a.Length;
sortUsingHash(a, n);
}
}
/* This code contributed by PrinciRaj1992 */
<script>
// javascript program to sort an array using hash
// function with negative values allowed.
function absolute(int x){
if(x<0) return (-1*x);
return x;
}
function sortUsingHash(a, n) {
// find the maximum element
var max = Math.max.apply(Math, a);
var min = absolute(Math.min.apply(Math, a));
// create a hash function upto the max size
var hashpos = Array(max).fill(0);
var hashneg = Array(min + 1).fill(0);
// traverse through all the elements and
// keep a count
for (i = 0; i < n; i++) {
if (a[i] >= 0)
hashpos[a[i]] += 1;
else
hashneg[absolute(a[i])] += 1;
}
// Traverse up to all negative elements and
// check if it is present or not. If it is
// present, then print the element the number
// of times it's present. Once we have printed
// n times, that means we have printed n elements
// so break out of the loop
for (i = min; i > 0; i--) {
if (hashneg[i] > 0) {
// print the element that number of times
// it's present. Print the negative element
for (j = 0; j < hashneg[i]; j++) {
document.write((-1) * i + " ");
}
}
}
// Traverse upto all elements and check if it is
// present or not. If it is present, then print
// the element the number of times it's present
// once we have printed n times, that means we
// have printed n elements, so break out of the
// loop
for (i = 0; i <= max; i++) {
// if present
if (hashpos[i] > 0) {
// print the element that number of times
// it's present
for (j = 0; j < hashpos[i]; j++) {
document.write(i + " ");
}
}
}
}
// Driver program to test the above function
var a = [ -1, -2, -3, -4, -5, -6, 8, 7, 5, 4, 3, 2, 1, 0 ];
var n = a.length;
sortUsingHash(a, n);
// This code contributed by Rajput-Ji
</script>
Output-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 7 8
Complexity:
Time complexity- The time complexity of this program is O(n + max), where n is the size of the input array and max is the maximum element in the array. This is because the program first finds the maximum element in the array using std::max_element, which takes O(n) time. It then creates two hash arrays of size max+1 and min+1, where min is the absolute value of the minimum element in the array. This takes O(max+min) time, but since min is always less than or equal to max, we can simplify this to O(max). The program then traverses through the input array once to fill in the hash arrays, which takes O(n) time. Finally, the program traverses through the two hash arrays, printing out the elements in sorted order. This takes O(max) time, since max is the size of the hash arrays. Therefore, the total time complexity of the program is O(n + max).
Space complexity -The space complexity of this program is O(max), since the program creates two hash arrays of size max+1 and min+1, where min is the absolute value of the minimum element in the array. However, since min is always less than or equal to max, we can simplify this to O(max). Therefore, the space complexity of the program is O(max).
Limitations:
- Can only sort array elements of limited range (typically from -10^6 to +10^6)
- Auxiliary space in worst cases is O(max_element) + O(min_element)
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