Created
May 13, 2013 08:59
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find the minimum and maximum element from an array with minimum time of comparision
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/** | |
* Time Complexity : Comparision time : 3 * n/2 + 1 | |
*/ | |
public int[] findMinMax(int[] A){ | |
int n = A.length; | |
if (n == 0) return null; | |
int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; | |
for (int i=0; i < n-1; i+=2){ | |
int smaller = A[i], bigger = A[i+1]; | |
if (A[i] > A[i+1){ | |
smaller = A[i+1]; | |
bigger = A[i]; | |
} | |
if (smaller < min) | |
min = smaller; | |
if (bigger > max) | |
max = bigger; | |
} | |
if (n % 2 == 1){ | |
if (A[n-1] < min) | |
min = A[n-1]; | |
else (A[n-1] > max) | |
max = A[n-1]; | |
} | |
return new int[]{min, max}; | |
} |
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