Difference between revisions of "TADM2E 4.8"
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Revision as of 18:14, 11 September 2014
(a): <pre> Sort S with any nlogn sorting method of your choice. for( int i = 1; i <= n; ++i ) {
int j = x  S[i]; Binary search for j in the subarray of S[i+1~n] and close the problem once it's been found;
} </pre> (b): <pre> Subtract each of S[1~n] from x to get a new array of real numbers T[1~n]. int i = 1, j = 1; while( i <=n && j <= n ) {
if( S[i] == T[j] ) { problem solved; break; } else { S[i] < T[j] ? ++i : ++j; }
} </pre>
Another Solution to part B without creating the additional Array.. <pre> i = 0; j = n  1;
for (i = 0; i < j; i++) {
while( (i < j) && (S[j] + S[i] > x ) { j; } if (x == (S[j] + S[i]) { return (S[i],S[j]); break; }
}
"i" scans from left to right, "j" from right to left, looking for the right pair... </pre>