Given an array A[1...N], it finds the position of the element with the minimum value between two given indices.
It returns the answer in constant time. The RMQ problem can be used to recover the lca (lowest common ancestor) in a graph in constant time. See http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor for more details.
Time complexity: O(N log(N))
Space complexity: O(N log(N))
Example
A = [-1 5 0 3 -6 4 2 1 0 -1];
N = length(A);
M = rmq(A);
% Find the position of the minimum element between indices i, j
i = 2;
j = 6;
k = floor(log2(j - i + 1));
if A(M(i,k+1)) <= A(M(j-2^k+1,k+1))
idx = M(i,k+1);
else
idx = M(j-2^k+1,k+1);
end