Created
May 25, 2015 12:05
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//I/P [1, 2, 4, ,6, 3, 7, 8] | |
//O/P 5 | |
//NOTE : array is sorted | |
//1 .total = n*(n+1)/2 leads to overflow if total can't be represented by the number O(n) | |
//2. XOR (given array) ^ (numbers from 1-n) = missing number O(n) | |
//3. Binary search (Find out the hole wheather in first half/second half) (O log n) | |
//Array: [1,2,3,4,5,6,8,9] | |
//Index: [0,1,2,3,4,5,6,7] | |
public static int binarySearchHelper(int a[],int start,int end) { | |
if (start < end) { | |
int mid = start + (end - start) / 2; | |
if (a[mid] - a[start] != (mid - start) ) { | |
//hole is in first half | |
if (mid - start == 1 && a[mid] - a[start] != 1) { | |
return mid; | |
} else { | |
return binarySearchHelper(a,start,mid-1); | |
} | |
} else if (a[end]-a[mid] != (end-mid) ) { | |
//hole is in second half | |
if (end - mid == 1 && a[end] - a[mid] != 1) { | |
return mid; | |
} else { | |
return binarySearchHelper(a,mid+1,end); | |
} | |
} else { | |
return -1; | |
} | |
} else { | |
return -1; | |
} | |
} | |
public static int findMissingNumber (int a[]) { | |
int len = a.length; | |
return binarySearchHelper (a,0,len); | |
} |
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