Minimum changes required to make two arrays identical
Last Updated :
02 Sep, 2022
Given two arrays, A and B with n elements each. The task is to make these two arrays identical i:e, for each 1\leq i \leq N , we want to make A_{i} = B_{i} . In a single operation, you can choose two integers x and y, and replace all the occurrences of x in both the arrays with y. Notice that regardless of the number of occurrences replaced, it will still be counted as a single operation. You have to output the minimum number of operations required.
Examples:
Input : 1 2 2
1 2 5
Output: 1
Here, (x, y) = (5, 2) hence ans = 1.
Input : 2 1 1 3 5
1 2 2 4 5
Output: 2
Here, (x, y) = (1, 2) and (3, 4) thus ans = 2.
Other pairs are also possible.
This problem can be solved with the help of Disjoint Set Union.
We will check all elements of both arrays i:e for each 1\leq i \leq N . If the elements belong to the same id then we skip it. Otherwise, we do a Union operation on both elements. Finally, the answer will be the sum of the sizes of all the different disjoint sets formed i:e ans = \sum_{i=1}^{N} (sz[i]-1) . We subtract 1 because, initially, we take the size of each set to be 1.
Implementation:
C++
// C++ program to find minimum changes
// required to make two arrays identical
#include <bits/stdc++.h>
using namespace std;
#define N 100010
/* 'id': stores parent of a node.
'sz': stores size of a DSU tree. */
int id[N], sz[N];
// Function to assign root
int Root(int idx)
{
int i = idx;
while (i != id[i])
id[i] = id[id[i]], i = id[i];
return i;
}
// Function to find Union
void Union(int a, int b)
{
int i = Root(a), j = Root(b);
if (i != j) {
if (sz[i] >= sz[j]) {
id[j] = i, sz[i] += sz[j];
sz[j] = 0;
}
else {
id[i] = j, sz[j] += sz[i];
sz[i] = 0;
}
}
}
// function to find minimum changes required
// to make both array equal.
int minChange(int n, int a[], int b[])
{
// Sets as single elements
for (int i = 0; i < N; i++)
id[i] = i, sz[i] = 1;
// Combine items if they belong to different
// sets.
for (int i = 0; i < n; ++i)
// true if both elements have different root
if (Root(a[i]) != Root(b[i]))
Union(a[i], b[i]); // make root equal
// Find sum sizes of all sets formed.
int ans = 0;
for (int i = 0; i < n; ++i)
if (id[i] == i)
ans += (sz[i] - 1);
return ans;
}
// Driver program
int main()
{
int a[] = { 2, 1, 1, 3, 5 }, b[] = { 1, 2, 2, 4, 5 };
int n = sizeof(a) / sizeof(a[0]);
cout << minChange(n, a, b);
return 0;
}
// Java program to find minimum changes
// required to make two arrays identical
class GFG{
static int N=100010;
/* 'id': stores parent of a node.
'sz': stores size of a DSU tree. */
static int[] id=new int[100010];
static int[] sz=new int[100010];
// Function to assign root
static int Root(int idx)
{
int i = idx;
while (i != id[i])
{
id[i] = id[id[i]];
i = id[i];
}
return i;
}
// Function to find Union
static void Union(int a, int b)
{
int i = Root(a);
int j = Root(b);
if (i != j) {
if (sz[i] >= sz[j]) {
id[j] = i;
sz[i] += sz[j];
sz[j] = 0;
}
else {
id[i] = j;
sz[j] += sz[i];
sz[i] = 0;
}
}
}
// function to find minimum changes required
// to make both array equal.
static int minChange(int n, int a[], int b[])
{
// Sets as single elements
for (int i = 0; i < N; i++)
{
id[i] = i;
sz[i] = 1;
}
// Combine items if they belong to different
// sets.
for (int i = 0; i < n; ++i)
// true if both elements have different root
if (Root(a[i]) != Root(b[i]))
Union(a[i], b[i]); // make root equal
// Find sum sizes of all sets formed.
int ans = 0;
for (int i = 0; i < n; ++i)
if (id[i] == i)
ans += (sz[i] - 1);
return ans;
}
// Driver program
public static void main(String[] args)
{
int a[] = { 2, 1, 1, 3, 5 }, b[] = { 1, 2, 2, 4, 5 };
int n = a.length;
System.out.println(minChange(n, a, b));
}
}
// This code is contributed by mits
# Python 3 program to find minimum changes
# required to make two arrays identical
N = 100010
# 'id':stores parent of a node
# 'sz':stores size of a DSU tree
ID = [0 for i in range(N)]
sz = [0 for i in range(N)]
# function to assign root
def Root(idx):
i = idx
while i != ID[i]:
ID[i], i = ID[ID[i]], ID[i]
return i
# Function to find Union
def Union(a, b):
i, j = Root(a), Root(b)
if i != j:
if sz[i] >= sz[j]:
ID[j] = i
sz[i] += sz[j]
sz[j] = 0
else:
ID[i] = j
sz[j] += sz[i]
sz[i] = 0
# function to find minimum changes
# required to make both array equal
def minChange(n, a, b):
# sets as single elements
for i in range(N):
ID[i] = i
sz[i] = 1
# Combine items if they belong
# to different sets
for i in range(n):
# true if both elements have
# different root
if Root(a[i]) != Root(b[i]):
Union(a[i], b[i])
# find sum sizes of all sets formed
ans = 0
for i in range(n):
if ID[i] == i:
ans += (sz[i] - 1)
return ans
# Driver Code
a = [2, 1, 1, 3, 5]
b = [1, 2, 2, 4, 5]
n = len(a)
print(minChange(n, a, b))
# This code is contributed
# by Mohit kumar 29 (IIIT gwalior)
// C# program to find minimum changes
// required to make two arrays identical
using System;
class GFG{
static int N=100010;
/* 'id': stores parent of a node.
'sz': stores size of a DSU tree. */
static int []id=new int[100010];
static int []sz=new int[100010];
// Function to assign root
static int Root(int idx)
{
int i = idx;
while (i != id[i])
{
id[i] = id[id[i]];
i = id[i];
}
return i;
}
// Function to find Union
static void Union(int a, int b)
{
int i = Root(a);
int j = Root(b);
if (i != j) {
if (sz[i] >= sz[j]) {
id[j] = i;
sz[i] += sz[j];
sz[j] = 0;
}
else {
id[i] = j;
sz[j] += sz[i];
sz[i] = 0;
}
}
}
// function to find minimum changes required
// to make both array equal.
static int minChange(int n, int []a, int []b)
{
// Sets as single elements
for (int i = 0; i < N; i++)
{
id[i] = i;
sz[i] = 1;
}
// Combine items if they belong to different
// sets.
for (int i = 0; i < n; ++i)
// true if both elements have different root
if (Root(a[i]) != Root(b[i]))
Union(a[i], b[i]); // make root equal
// Find sum sizes of all sets formed.
int ans = 0;
for (int i = 0; i < n; ++i)
if (id[i] == i)
ans += (sz[i] - 1);
return ans;
}
// Driver program
public static void Main()
{
int []a = { 2, 1, 1, 3, 5 };
int []b = { 1, 2, 2, 4, 5 };
int n = a.Length;
Console.WriteLine(minChange(n, a, b));
}
}
// This code is contributed by anuj_67..
<?php
// PHP program to find minimum changes
// required to make two arrays identical
$N = 100010;
/* 'id': stores parent of a node.
'sz': stores size of a DSU tree. */
$id = array_fill(0, $N, NULL);
$sz = array_fill(0, $N, NULL);
// Function to assign root
function Root($idx)
{
global $id;
$i = $idx;
while ($i != $id[$i])
{
$id[$i] = $id[$id[$i]];
$i = $id[$i];
}
return $i;
}
// Function to find Union
function Union($a, $b)
{
global $sz, $id;
$i = Root($a);
$j = Root($b);
if ($i != $j)
{
if ($sz[$i] >= $sz[$j])
{
$id[$j] = $i;
$sz[$i] += $sz[$j];
$sz[$j] = 0;
}
else
{
$id[$i] = $j;
$sz[$j] += $sz[$i];
$sz[$i] = 0;
}
}
}
// function to find minimum changes
// required to make both array equal.
function minChange($n, &$a, &$b)
{
global $id, $sz, $N;
// Sets as single elements
for ($i = 0; $i < $N; $i++)
{
$id[$i] = $i;
$sz[$i] = 1;
}
// Combine items if they belong to
// different sets.
for ($i = 0; $i < $n; ++$i)
// true if both elements have
// different roots
if (Root($a[$i]) != Root($b[$i]))
Union($a[$i], $b[$i]); // make root equal
// Find sum sizes of all sets formed.
$ans = 0;
for ($i = 0; $i < $n; ++$i)
if ($id[$i] == $i)
$ans += ($sz[$i] - 1);
return $ans;
}
// Driver Code
$a = array(2, 1, 1, 3, 5);
$b = array(1, 2, 2, 4, 5);
$n = sizeof($a);
echo minChange($n, $a, $b);
// This code is contributed by ita_c
?>
<script>
// Javascript program to find minimum changes
// required to make two arrays identical
let N=100010;
/* 'id': stores parent of a node.
'sz': stores size of a DSU tree. */
let id=new Array(100010);
let sz=new Array(100010);
// Function to assign root
function Root(idx)
{
let i = idx;
while (i != id[i])
{
id[i] = id[id[i]];
i = id[i];
}
return i;
}
// Function to find Union
function Union(a,b)
{
let i = Root(a);
let j = Root(b);
if (i != j) {
if (sz[i] >= sz[j]) {
id[j] = i;
sz[i] += sz[j];
sz[j] = 0;
}
else {
id[i] = j;
sz[j] += sz[i];
sz[i] = 0;
}
}
}
// function to find minimum changes required
// to make both array equal.
function minChange(n,a,b)
{
// Sets as single elements
for (let i = 0; i < N; i++)
{
id[i] = i;
sz[i] = 1;
}
// Combine items if they belong to different
// sets.
for (let i = 0; i < n; ++i)
// true if both elements have different root
if (Root(a[i]) != Root(b[i]))
Union(a[i], b[i]); // make root equal
// Find sum sizes of all sets formed.
let ans = 0;
for (let i = 0; i < n; ++i)
if (id[i] == i)
ans += (sz[i] - 1);
return ans;
}
// Driver program
let a=[2, 1, 1, 3, 5 ];
let b=[ 1, 2, 2, 4, 5 ];
let n = a.length;
document.write(minChange(n, a, b));
// This code is contributed by rag2127
</script>
Time Complexity: O(N + n) where N is the maximum possible value of an array item and n is the number of elements in the array.
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