Difference between revisions of "TADM2E 4.34"
From Algorithm Wiki
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<pre> | <pre> | ||
FindMinIntMissing(A): | FindMinIntMissing(A): | ||
− | // elements in the array are continuous, the smallest missing integer | + | // Trivial case 1 |
− | if(A[n - 1] - A[0] == n - 1) | + | if(A[0] > 1) |
+ | return 1; | ||
+ | // Trivial case 2: elements in the array are continuous from 1 to n, the smallest missing integer must n + 1 | ||
+ | else if(A[n - 1] - A[0] == n - 1) | ||
{ | { | ||
− | + | return n + 1; | |
− | |||
− | |||
− | |||
} | } | ||
// there is a gap in A, find the gap that is smallest | // there is a gap in A, find the gap that is smallest |
Revision as of 00:24, 23 September 2016
Assume array indices are zero based, use modified binary search to find the smallest integer missing:
FindMinIntMissing(A): // Trivial case 1 if(A[0] > 1) return 1; // Trivial case 2: elements in the array are continuous from 1 to n, the smallest missing integer must n + 1 else if(A[n - 1] - A[0] == n - 1) { return n + 1; } // there is a gap in A, find the gap that is smallest else { return findGap(A, 0, n - 1); } FindGap(A, first, last): // only two elements left, the smallest gap must be the integer right after A[first] if(last == first + 1) return A[first] + 1; // recursively looking for the partition containing the smallest gap else mid = (first + last) / 2; // first half doesn't have gap, looking for gap in the second half if(A[mid] - A[first] == mid - first) return findGap(A, mid, last); // looking for the gap in the first half else return findGap(A, first, mid);