Google Interview Question for Java Developers


Country: United States




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0
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You need to swap 1 by 1, 2 by 2, 4 by 4, 8 by 8 and so on.
If you use shifting strategy it take O(n^2), using this strategy it takes O(n log n).
a0 a1 a2 a3 a4 a5 a6 a7 b0 b1 b2 b3 b4 b5 b6 b7
(a0) (a1) (a2) (a3) (a4) (a5) (a6) (a7) (b0) (b1) (b2) (b3) (b4) (b5) (b6) (b7)
(a0 b0) (a2 b2) (a4 b4) (a6 b6) (a1 b1) (a3 b3) (a5 b5) (a7 b7)
(a0 b0 a1 b1) (a4 b4 a5 b5) (a2 b2 a3 b3) (a6 b6 a7 b7)
(a0 b0 a1 b1 a2 b2 a3 b3) (a4 b4 a5 b5 a6 b6 a7 b7)
(a0 b0 a1 b1 a2 b2 a3 b3 a4 b4 a5 b5 a6 b6 a7 b7)

- Nooreddin February 01, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

You need to swap 1 by 1, 2 by 2, 4 by 4, 8 by 8 and so on.
If you use shifting strategy it take O(n^2), using this strategy it takes O(n log n).
a0 a1 a2 a3 a4 a5 a6 a7 b0 b1 b2 b3 b4 b5 b6 b7
(a0) (a1) (a2) (a3) (a4) (a5) (a6) (a7) (b0) (b1) (b2) (b3) (b4) (b5) (b6) (b7)
(a0 b0) (a2 b2) (a4 b4) (a6 b6) (a1 b1) (a3 b3) (a5 b5) (a7 b7)
(a0 b0 a1 b1) (a4 b4 a5 b5) (a2 b2 a3 b3) (a6 b6 a7 b7)
(a0 b0 a1 b1 a2 b2 a3 b3) (a4 b4 a5 b5 a6 b6 a7 b7)
(a0 b0 a1 b1 a2 b2 a3 b3 a4 b4 a5 b5 a6 b6 a7 b7)

- nooreddin February 01, 2018 | Flag Reply
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0
of 0 vote

public static void changeArray(String[] arr){
		if(arr.length ==0)
			return;
		int index = arr.length / 2;
		while(index < arr.length){
			String temp = arr[index];
			int counter = index -1;
			while(counter >= 0 && LessThan(arr[counter],temp)){
				arr[counter+1] = arr[counter];
				counter--;
			}
			arr[counter+1] = temp;
			index++;
		}
	}
	private static boolean LessThan(String a, String b){
		return Integer.valueOf(b.charAt(1)) < Integer.valueOf(a.charAt(1));
	}

- Anonymous February 01, 2018 | Flag Reply
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0
of 0 vote

public static void changeArray(String[] arr){
		if(arr.length ==0)
			return;
		int index = arr.length / 2;
		while(index < arr.length){
			String temp = arr[index];
			int counter = index -1;
			while(counter >= 0 && LessThan(arr[counter],temp)){
				arr[counter+1] = arr[counter];
				counter--;
			}
			arr[counter+1] = temp;
			index++;
		}
	}
	private static boolean LessThan(String a, String b){
		return Integer.valueOf(b.charAt(1)) < Integer.valueOf(a.charAt(1));
	}

- Tarun February 01, 2018 | Flag Reply
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0
of 0 vote

public static void rearrange(int[] arr) {
    if (arr == null || arr.length % 2 == 1) return;

    int currIdx = (arr.length - 1) / 2;
    while (currIdx > 0) {
        int count = currIdx, swapIdx = currIdx;

        while (count-- > 0) {
            int temp = arr[swapIdx + 1];
            arr[swapIdx + 1] = arr[swapIdx];
            arr[swapIdx] = temp;
            swapIdx++;
        }

        currIdx--;
    }
}

- jkgriesser February 02, 2018 | Flag Reply
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0
of 0 vote

O(n) solution would be something as follows:
for each even element a_i in a, do the following:
index=i
while index <n : index=index*2-1
swap (a_index, b_(i/2))
while index !=i : swap (a_index,a_((index-1)/2)); index=(index+1)/2
Repeat for array b(1..n/2) b(n/2+1)
Total operations: n+n/2+n/4...=O(n)

- jayz February 06, 2018 | Flag Reply
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0
of 0 vote

>>> sl
['a0', 'a1', 'a2', 'b0', 'b1', 'b2']

>>> sl.sort(key=lambda x: x[1])

>>> sl
['a0', 'b0', 'a1', 'b1', 'a2', 'b2']

- David Dhas July 28, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

#include<stdio.h>
#include<conio.h>
int main()
{
}

- Kshitij sahu February 01, 2018 | Flag Reply


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