Bear and Steady Gene
Problem Statement :
A gene is represented as a string of length (where is divisible by ), composed of the letters , , , and . It is considered to be steady if each of the four letters occurs exactly times. For example, and are both steady genes. Bear Limak is a famous biotechnology scientist who specializes in modifying bear DNA to make it steady. Right now, he is examining a gene represented as a string . It is not necessarily steady. Fortunately, Limak can choose one (maybe empty) substring of and replace it with any string of the same length. Modifying a large substring of bear genes can be dangerous. Given a string , can you help Limak find the length of the smallest possible substring that he can replace to make a steady gene? Note: A substring of a string is a subsequence made up of zero or more contiguous characters of . As an example, consider . The substring just before or after can be replaced with or . One selection would create . Function Description Complete the function in the editor below. It should return an integer that represents the length of the smallest substring to replace. steadyGene has the following parameter: gene: a string Input Format The first line contains an interger divisible by , that denotes the length of a string . The second line contains a string of length n. Output Format Print the length of the minimum length substring that can be replaced to make gene stable.
Solution :
Solution in C :
In C++ :
#include <bits/stdc++.h>
#define F first
#define S second
#define X real()
#define Y imag()
using namespace std;
typedef long long ll;
typedef long double ld;
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
string s;
cin>>n>>s;
int a=0;
int c=0;
int t=0;
int g=0;
for (int i=0;i<n;i++){
if (s[i]=='A') a++;
if (s[i]=='C') c++;
if (s[i]=='T') t++;
if (s[i]=='G') g++;
}
a-=n/4;
c-=n/4;
t-=n/4;
g-=n/4;
int i2=0;
int ca=0;
int cc=0;
int ct=0;
int cg=0;
int v=n;
for (int i=0;i<n;i++){
while (i2<n&&(ca<a||cc<c||ct<t||cg<g)){
if (s[i2]=='A') ca++;
if (s[i2]=='C') cc++;
if (s[i2]=='T') ct++;
if (s[i2]=='G') cg++;
i2++;
}
if (!(ca<a||cc<c||ct<t||cg<g)){
v=min(v, i2-i);
}
if (s[i]=='A') ca--;
if (s[i]=='C') cc--;
if (s[i]=='T') ct--;
if (s[i]=='G') cg--;
}
cout<<v<<endl;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static int code(char c) {
if (c == 'A') return 0;
if (c == 'C') return 1;
if (c == 'G') return 2;
return 3;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
in.nextLine();
String str = in.nextLine();
int j = n-1;
int[] count = new int[4];
while (true) {
if (j == -1) {
System.out.println("0");
return;
}
if (count[code(str.charAt(j))] + 1 > n/4) {
j++;
break;
}
count[code(str.charAt(j))]++;
j--;
}
int result = j;
for (int i = 0; i < n; i++) {
count[code(str.charAt(i))]++;
while(count[code(str.charAt(i))] > n/4) {
if (j == n) {
System.out.println(result);
return;
}
count[code(str.charAt(j))]--;
j++;
}
result = Math.min(result, j - i - 1);
}
System.out.println(result);
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>
#define MAX_L 500002
static int cvt(int c)
{
switch (c) {
case 'A':
return 0;
case 'T':
return 1;
case 'C':
return 2;
case 'G':
return 3;
}
abort();
}
static bool check(int cnts[][MAX_L + 1], int n, int i, int l)
{
bool ok = 1;
int sum = 0;
for (int j = 0; j < 4; j++) {
int diff = n/4 - (cnts[j][n] - (cnts[j][i + l] - cnts[j][i]));
ok &= diff >= 0;
sum += diff;
}
return ok && sum == l;
}
static int subst(char s[MAX_L], int n)
{
int res = INT_MAX;
static int cnts[4][MAX_L + 1];
for (int i = 1; i <= n; i++) {
for (int j = 0; j < 4; j++) {
cnts[j][i] = cnts[j][i-1];
}
cnts[s[i-1]][i] += 1;
}
for (int i = 0; i < n; i++) {
int len = 1;
int l, r;
while (i + len <= n) {
if (check(cnts, n, i, len))
break;
len *= 2;
}
if (i + len > n) {
len = n - i;
if (!check(cnts, n, i, len))
continue;
}
l = len / 2;
r = len;
while (l < r) {
int m = l + (r - l)/2;
if (check(cnts, n, i, m)) {
r = m;
len = r;
} else {
l = m + 1;
}
}
if (len < res)
res = len;
}
return res;
}
int main()
{
int n;
static char s[MAX_L];
fgets(s, MAX_L, stdin);
sscanf(s, "%d", &n);
fgets(s, MAX_L, stdin);
for (int i = 0; i < n; i++) {
s[i] = cvt(s[i]);
}
printf("%d", subst(s, n));
return 0;
}
In Python3 :
from collections import Counter
n = int(input())
m = n // 4
s = input()
cnt = {'G': 0, 'A': 0, 'C': 0, 'T': 0}
cnt.update(Counter(s))
if all(map(lambda x: x == m, cnt.values())):
print(0)
exit()
low, high = 0, n
while high - low > 1:
mid = (high + low) // 2
cnt_tmp = dict(cnt)
for i in range(mid):
cnt_tmp[s[i]] -= 1
if all(map(lambda x: x <= m, cnt_tmp.values())):
high = mid
continue
for i in range(mid, n):
cnt_tmp[s[i - mid]] += 1
cnt_tmp[s[i]] -= 1
if all(map(lambda x: x <= m, cnt_tmp.values())):
high = mid
break
else:
low = mid
print(high)
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