how can we find lexicographically smallest anagram of string
// C++ program to convert string X to
// string Y which minimum number of changes.
#include <bits/stdc++.h>
using namespace std;
#define MAX 26
// Function that converts string X
// into lexicographically smallest
// anagram of string Y with minimal changes
void printAnagramAndChanges(string X, string Y)
{
int countx[MAX] = {0}, county[MAX] = {0},
ctrx[MAX] = {0}, ctry[MAX] = {0};
int change = 0;
int l = X.length();
// Counting frequency of characters
// in each string.
for (int i = 0; i < l; i++) {
countx[X[i] - 'A']++;
county[Y[i] - 'A']++;
}
// We maintain two more counter arrays
// ctrx[] and ctry[]
// Ctrx[] maintains the count of extra
// elements present in string X than
// string Y
// Ctry[] maintains the count of
// characters missing from string X
// which should be present in string Y.
for (int i = 0; i < MAX; i++) {
if (countx[i] > county[i])
ctrx[i] += (countx[i] - county[i]);
else if (countx[i] < county[i])
ctry[i] += (county[i] - countx[i]);
change += abs(county[i] - countx[i]);
}
for (int i = 0; i < l; i++) {
// This means that we cannot edit the
// current character as it's frequency
// in string X is equal to or less
// than the frequency in string Y.
// Thus, we go to the next position
if (ctrx[X[i] - 'A'] == 0)
continue;
// Here, we try to find that character,
// which has more frequency in string Y
// and less in string X. We try to find
// this character in lexicographical
// order so that we get
// lexicographically smaller string
int j;
for (j = 0; j < MAX; j++)
if ((ctry[j]) > 0)
break;
// This portion deals with the
// lexicographical property.
// Now, we put a character in string X
// when either this character has smaller
// value than the character present there
// right now or if this is the last position
// for it to exchange, else we fix the
// character already present here in
// this position.
if (countx[X[i] - 'A'] == ctrx[X[i] - 'A']
|| X[i] - 'A' > j) {
countx[X[i] - 'A']--;
ctrx[X[i] - 'A']--;
ctry[j]--;
X[i] = 'A' + j;
}
else
countx[X[i] - 'A']--;
}
cout << "Anagram : " << X << endl;
cout << "Number of changes made : " << change / 2;
}
// Driver program
int main()
{
string x = "CDBABC", y = "ADCABD";
printAnagramAndChanges(x, y);
return 0;
}
