Hi,
This one is quite easy. You have to calculate the number of one's before
every zero and add them up. That's it.
public class Test1 {
public void printArray(int[] tmpArr) {
for (int i : tmpArr) {
System.out.println(i);
}
}
public int calculateMinSwaps(int[] tmpArr) {
int minSwaps = 0;
int numberOfOne = 0;
for (int i : tmpArr) {
if (i == 1) {
numberOfOne++;
} else {
minSwaps += numberOfOne;
}
}
return minSwaps;
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Test1 test1 = new Test1();
int[] minSwaps = {
1,1,1,0,0,0,1,0
};
// test1.printArray(minSwaps);
int minswap = test1.calculateMinSwaps(minSwaps);
System.out.println(minswap);
}
}
On Saturday, June 23, 2012 11:34:55 AM UTC+5:30, zerocool142 wrote:
>
> Given an array containing sequence of bits (0 or 1), you have to sort
> this array in the ascending order i.e. all 0' in first part of array
> followed by all 1's. The constraints is that you can swap only the
> adjacent elements in the array. Find the minimum number of swaps
> required to sort the given input array.
>
> Example: Given the array (0,0,1,0,1,0,1,1) the minimum number of swaps
> is 3.
>
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