him4211
BAN USER- 0of 0 votes
AnswersYou have given a string which consists of four characters {a,c,t,g}, find minimum length of unique substring which occurs only once.
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Eg:
Input :
#1 aacc
#2 actg
Output:
#1 2 3
#2 1 4
Explain:
aacc -> {aa, ac, cc} are the smallest unique string with single occurrence.| Report Duplicate | Flag | PURGE
Samsung Software Developer - 0of 0 votes
AnswerHow to divide a circular array into k group of contiguous element such that difference between maximum sum and minimum sum is minimum. Each group have contiguous element of array. For e.g If the array is as follow. [6 13 10 2] and k=2 then o/p should be 18(6+10+2)-13=5. As array is circular 6,10,2 are contiguous element of array.
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For e.g If the array is as follow. [6 13 2 10] and k=2 then o/p should be 16(6+10)-15(13+2)=1. As array is circular 6,10 are contiguous element of array.
For e.g If the array is as follow. [100 92 133 201 34 34 34 94 108] and k=4 then group as follow 208(108,100), 225(92,133), (201), 196(34,34,34,94) so 225-196=29| Report Duplicate | Flag | PURGE
Samsung Software Developer Algorithm
package general;
import java.util.Arrays;
import java.util.TreeMap;
public class SieveOfEratosthenese {
public static void getPrimesBySeieveOfEraosthenes(int n, boolean primes[]) {
for(int i=2; i<= Math.sqrt(n); i++){
if(primes[i]){
for(int j=i*i;j<=n; j+=i){
primes[j] = false;
}
}
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int n = 16;
boolean[] primes = new boolean[n+1];
Arrays.fill(primes, true);
getPrimesBySeieveOfEraosthenes(n, primes);
for (int i = 2; i <= n; i++) {
if (primes[i])
System.out.print((i) + " ");
}
TreeMap <Integer, Integer> sums = new TreeMap<>();
int sum =0;
for (int i = 2; i <= n; i++) {
if (primes[i]){
sums.put(i-1, sum);
sum +=i;
}
}
if(primes[n]){
sums.put(n,sum);
}
System.out.println( "\n"+ sums.get(sums.ceilingKey(4)));
}
}
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- him4211 April 22, 2019