## Google Interview Question

Software Engineers**Country:**United States

package findSubArray;

import java.util.*;

public class findSubArray {

public static int findsubArray(int arr[], int k) {

int n = arr.length, i, j;

int array[][] = new int[n][n];

for(i=0;i<n;i++) {

array[0][i] = arr[i];

}

for(i=0;i<n;i++) {

array[i][0] = arr[0];

}

for(i=1; i<n; i++) {

for(j=i;j<n;j++) {

array[i][j] = arr[j] * array[i-1][j-1];

System.out.println("i: "+i+" j: "+j+" arr[i-1][j-1]: "+array[i-1][j-1]+" arr[i][j]: "+array[i][j]);

if(array[i][j] % k ==0) {

if(i != 0) {

System.out.println("End of subarray is: "+arr[j]);

return i+1;

}

}

}

}

return 0;

}

public static void main(String[] args) {

// TODO Auto-generated method stub

int array[] = {1, 9, 16, 5, 4, 3, 2};

int k = 720;

int answer = findsubArray(array, k);

System.out.println("Minimum lenth of subarray is: "+answer);

}

}

If those the number of those lamps fits into an integer of some sort (or really, this requirement could be just the ranges you flip), I think this solution could work too (not tested):

```
int get_range(unsigned long n, int i, int j) {
return (n >> i) & ((1 << j) - 1);
}
int flip(unsigned long n, int i, int j) {
return n + (~get_range(n, i, j) - get_range(n, i, j));
}
int isOn(unsigned long n, int i) {
return n & (1 << i);
}
```

Space O(N), Update O(1) if the bitwise operators are O(1) as expected, query O(1)

```
hasSumMultK :: (Map.Map Int Int) -> Int ->[Int] -> Int-> Bool
hasSumMultK m s [] _ = Map.member s m
hasSumMultK m sum (x:xs) k = let newSum = (sum + x) `mod` k
in (Map.member sum m) ||
hasSumMultK (Map.insert sum sum m) newSum xs k
hasSumMultOfK :: [Int] -> Int -> Bool
hasSumMultOfK = hasSumMultK Map.empty 0
```

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