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Rearrange String k Distance Apart in C++
Suppose we have a non-empty string s and an integer k; we have to rearrange the string such that the same characters are at least distance k from each other. Given strings are in lowercase letters. If there is no way to rearrange the strings, then we will an empty string.
So, if the input is like s = "aabbcc", k = 3, then the output will be "abcabc" this is because same letters are at least distance 3 from each other.
To solve this, we will follow these steps −
ret := an empty string
Define one map m
n := size of s
for initialize i := 0, when i < n, update (increase i by 1), do −
(increase m[s[i]] by 1)
Define one priority queue pq
for each key-value pair it in m, do −
temp := a pair with key and value of it
insert temp into pq
(increase it by 1)
Define one deque dq
while (not pq is empty), do −
curr := top of pq
delete element from pq
ret := ret + curr.first
(decrease curr.second by 1)
insert curr at the end of dq
if size of dq >= k, then −
curr := first element of dq
delete front element from dq
if curr.second > 0, then −
insert curr into pq
while (not dq is empty and first element of dq is same as 0), do −
delete front element from dq
return (if dq is empty, then ret, otherwise blank string)
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
struct Comparator {
bool operator()(pair<char, int> a, pair<char, int> b) {
return !(a.second > b.second);
}
};
class Solution {
public:
string rearrangeString(string s, int k) {
string ret = "";
unordered_map<char, int> m;
int n = s.size();
for (int i = 0; i < n; i++) {
m[s[i]]++;
}
unordered_map<char, int>::iterator it = m.begin();
priority_queue<pair<char, int<, vector<pair<char, int>>,
Comparator> pq;
while (it != m.end()) {
pair<char, int> temp = {it->first, it->second};
pq.push(temp);
it++;
}
deque<pair<char, int>> dq;
while (!pq.empty()) {
pair<char, int> curr = pq.top();
pq.pop();
ret += curr.first;
curr.second--;
dq.push_back(curr);
// cout << ret << " " << dq.size() << endl;
if (dq.size() >= k) {
curr = dq.front();
dq.pop_front();
if (curr.second > 0)
pq.push(curr);
}
}
while (!dq.empty() && dq.front().second == 0)
dq.pop_front();
return dq.empty() ? ret : "";
}
};
main() {
Solution ob;
cout << (ob.rearrangeString("aabbcc", 3));
}Input
"aabbcc", 3
Output
bacbac
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