Election in Sylhet City

• For each number, find out the range where it is the largest
• For each number in left part, find out what least amount is needed in right part to surpass the max value of this part and store right part {start point, end point}
• Or do vice-versa in case right part has less elements
• There can be as much as $n(lg(n))$ such pairs
• Now build a segment tree and assume all the numbers are dead at the beginning
• Keep making numbers alive in descending order
• Use stored {start point, end point}s to make query when all the numbers more or equal needed by respective {start point, end point} are added to the segment tree
• For each segment, the max value can pair up with a number from the left or right part as well. Take care of that case.

Complexity can go up to $n{(lg(n))}^2$.