Problem Link - Devious Minister
Problem Statement
The government of Siruseri has just commissioned one of the longest and most modern railway routes in the world. This route runs the entire length of Siruseri and passes through many of the big cities and a large number of small towns and villages in Siruseri.
The railway stations along this route have all been constructed keeping in mind the comfort of the travellers. Every station has big parking lots, comfortable waiting rooms and plenty of space for eateries. The railway authorities would like to contract out the catering services of these eateries.
The Siruseri Economic Survey has done a through feasibility study of the different stations and documented the expected profits (or losses) for the eateries in all the railway stations on this route. The authorities would like to ensure that every station is catered to. To prevent caterers from bidding only for profitable stations, the authorities have decided to give out catering contracts for contiguous segments of stations.
The minister in charge realises that one of the bidders is his bitter adversary and he has decided to hand out as useless a segment as possible to him. On the other hand, he does not want to be seen to be blatantly unfair by handing out a large loss-making section to the adversary. Instead he wants to find the largest segment whose sum is closest to 0, so that his adversary spends all his time running a large number of canteens and makes either a small loss or a small profit or, even better, nothing at all!
In other words, if the profits/losses at the stations are p[1],p[2],...,p[N] the minister would like to handover a sequence i,i+1,...,j such that the absolute value of p[i]+p[i+1]+...+p[j] is minimized. If there is more than one sequence with this minimum absolute value then he would like to hand over the longest one.
Approach
To solve this problem, we aim to find a subarray whose sum
is closest to zero
. We use a prefix sum array to efficiently compute the sums
of all contiguous segments. By sorting the prefix sums and their indices, we can identify pairs of prefix sums that are closest in value, as their difference represents the sum of a subarray. For every pair of sorted prefix sums, the absolute difference between the sums is calculated, and the segment is recorded if it has a smaller absolute sum than the current best. If there is a tie, the longer segment is preferred. Finally, the segment’s endpoints and the corresponding sum are determined based on the indices of the closest prefix sums.
Time Complexity
Sorting the prefix sums takes O(n \log n), and iterating through the array involves O(n). Thus, the overall time complexity is O(n \log n).
Space Complexity
The algorithm requires O(n) extra space for the prefix sum and index arrays.