Given a string of ‘0’, ‘1’ and ‘2’. The task is to find the minimum replacements in the string such that the differences between the indexes of the same characters is divisible by 3.
Examples:
Input: s = “2101200”
Output: 3
1201201 or 2102102 can be the resultant string
which has 3 replacements.Input: s = “012”
Output: 0
Approach: There can be 6 different strings such that the difference between the index of similar characters is divisible by 3. Hence generate all 6 different strings, and compare the replacements done with the original string. Store the string which has a minimal number of replacements. The different strings can be generated using next_permutation in C++.
Below is the implementation of the above approach:
// C++ program to find minimum replacements // such that the difference between the // index of the same characters // is divisible by 3 #include <bits/stdc++.h> using namespace std;
// Function to count the number of // minimal replacements int countMinimalReplacements(string s)
{ int n = s.length();
int mini = INT_MAX;
string dup = "012" ;
// Generate all permutations
do {
// Count the replacements
int dif = 0;
for ( int i = 0; i < n; i++)
if (s[i] != dup[i % 3])
dif++;
mini = min(mini, dif);
} while (next_permutation(dup.begin(), dup.end()));
// Return the replacements
return mini;
} // Driver Code int main()
{ string s = "2101200" ;
cout << countMinimalReplacements(s);
return 0;
} |
// Java program to find minimum replacements // such that the difference between the // index of the same characters // is divisible by 3 class GFG {
// Function to count the number of
// minimal replacements
static int countMinimalReplacements(String s)
{
int n = s.length();
int mini = Integer.MAX_VALUE;
char [] dup = "012" .toCharArray();
// Generate all permutations
do {
// Count the replacements
int dif = 0 ;
for ( int i = 0 ; i < n; i++) {
if (s.charAt(i) != dup[i % 3 ]) {
dif++;
}
}
mini = Math.min(mini, dif);
} while (next_permutation(dup));
// Return the replacements
return mini;
}
static boolean next_permutation( char [] p)
{
for ( int a = p.length - 2 ; a >= 0 ; --a) {
if (p[a] < p[a + 1 ]) {
for ( int b = p.length - 1 ;; --b) {
if (p[b] > p[a]) {
char t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1 ; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true ;
}
}
}
}
return false ;
}
// Driver Code
public static void main(String args[])
{
String s = "2101200" ;
System.out.println(countMinimalReplacements(s));
}
} /* This code contributed by PrinciRaj1992 */ |
# Python program to find minimum replacements # such that the difference between the # index of the same characters # is divisible by 3 import sys
# Function to count the number of # minimal replacements def countMinimalReplacements(s):
n = len (s)
mini = sys.maxsize
dup = "012"
# Generate all permutations
for i in range ( 6 ):
# Count the replacements
dif = 0
for j in range (n):
if s[j] ! = dup[j % 3 ]:
dif + = 1
mini = min (mini, dif)
dup = dup[ 1 :] + dup[ 0 ]
# Return the replacements
return mini
# Driver Code if __name__ = = '__main__' :
s = "2101200"
print (countMinimalReplacements(s))
|
// C# program to find minimum replacements // such that the difference between the // index of the same characters // is divisible by 3 using System;
class GFG {
// Function to count the number of
// minimal replacements
static int countMinimalReplacements(String s)
{
int n = s.Length;
int mini = int .MaxValue;
char [] dup = "012" .ToCharArray();
// Generate all permutations
do {
// Count the replacements
int dif = 0;
for ( int i = 0; i < n; i++) {
if (s[i] != dup[i % 3]) {
dif++;
}
}
mini = Math.Min(mini, dif);
} while (next_permutation(dup));
// Return the replacements
return mini;
}
static bool next_permutation( char [] p)
{
for ( int a = p.Length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for ( int b = p.Length - 1;; --b) {
if (p[b] > p[a]) {
char t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.Length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true ;
}
}
}
}
return false ;
}
// Driver Code
public static void Main(String[] args)
{
String s = "2101200" ;
Console.WriteLine(countMinimalReplacements(s));
}
} // This code has been contributed by 29AjayKumar |
<script> // JavaScript program to find minimum replacements
// such that the difference between the
// index of the same characters
// is divisible by 3
// Function to count the number of
// minimal replacements
function countMinimalReplacements(s) {
var n = s.length;
//Max Integer Value
var mini = 2147483647;
var dup = "012" .split( "" );
// Generate all permutations
do {
// Count the replacements
var dif = 0;
for ( var i = 0; i < n; i++) {
if (s[i] !== dup[i % 3]) {
dif++;
}
}
mini = Math.min(mini, dif);
} while (next_permutation(dup));
// Return the replacements
return mini;
}
function next_permutation(p) {
for ( var a = p.length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for ( var b = p.length - 1; ; --b) {
if (p[b] > p[a]) {
var t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true ;
}
}
}
}
return false ;
}
// Driver Code
var s = "2101200" ;
document.write(countMinimalReplacements(s));
</script>
|
3
Complexity Analysis:
- Time Complexity: O(3*N), as we are a loop to traverse N times. Where n is the length of the string.
- Auxiliary Space: O(1), as we are not using any extra space.