# Smallest String With Swaps

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Problem statement

You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string.

You can swap the characters at any pair of indices in the given pairs any number of times.

Return the lexicographically smallest string that s can be changed to after using the swaps.

Another solution with disjoint sets. This problem is not just forming disjoint sets. Tricky logic to sort each disjoint set.

Below is the solution with descriptive comments:

int find(int *rt, int x){
int rx = rt[x];
if (rx == x)
return x;
return rt[x] = find(rt, rx);
}
void union_link(int *rt, int *rnk, int x, int y){
int rx = find(rt, x);
int ry = find(rt, y);
if (rx !=ry){
if (rnk[rx] > rnk[ry])
rt[ry] =rx;
else if (rnk[ry] > rnk[rx])
rt[rx] = ry;
else {
rt[ry] = rx;
rnk[rx] +=1;
}
}
}
/* djset contains array of 26 elements. Each array element corresponds to the count of each lower alphabet.
each disjoint set can use this structure instance.
each disjoint set can have lower alphabets a-z.
*/
struct djset {
int chcnt[26];
char ind;
};
char * smallestStringWithSwaps(char * s, int** pairs, int pairsSize, int* pairsColSize){
int n = strlen(s);

int rt[n];
/* frame disjoint set */
int ii, jj;
{
int rnk[n];
for (ii=0; ii< n; ii++){
rt[ii] = ii;
rnk[ii] = 1;
}
}

int numdjsets=0;
int ind_djsets[n];
/* get num of disjoint sets */
for (ii=0; ii< n; ii++){
jj = find(rt, ii);
if (jj == ii) {
ind_djsets[ii] = numdjsets;
numdjsets++;
}
}

/* Define a djset for each disjoint set.
Max number of disjoint sets can be of string length
*/
/* define array indices of djsets */
struct djset dsets[numdjsets+1];

memset(dsets, 0, sizeof(dsets));

for (ii=0; ii<n; ii++){
/* get the root element corresponding to each alphabet */
jj = rt[ii];

jj = ind_djsets[jj];
/* increment the corresponding alphabet count in root djset */
dsets[jj].chcnt[s[ii]-'a'] += 1;
}

/* keep the alphabets of each disjoint set in ascending order
Get the count of first non-zero elements in alphabets array,
assign it, reduce , then go for next non-zero element
*/
for (ii=0; ii<n; ii++){
/* get the root element */
jj = rt[ii];
jj = ind_djsets[jj];
/* get the lowest existing alphabet index of disjoint set */
while (!dsets[jj].chcnt[dsets[jj].ind]) dsets[jj].ind++;

/* assign the lowest existing alphabet of disjoint set element s[ii] */
s[ii] = 'a' + dsets[jj].ind;

/* go for the next lowest alphabet
by reducing the count of existing alphabet,
in loop, it goes to next alphabet count in next occurrence */
dsets[jj].chcnt[dsets[jj].ind] -= 1;
}

/* above loop ensures to arrange alphabets in ascending order */
return s;
}