The Problem:
There is a one-dimensional road of infinite length, which extends from -∞ to+ ∞. There are "n' one-dimensional transistors placed on the road at various points. Each transistor emits a signal. For every transistor, the signal can be transmitted across a region "r" which may differ from transmitter to transmitter. Since the road is one-dimensional, by radius "r", we mean that the signal can be received up to "r" distance to its left and "r" distance to its right.
There is a group of one-dimensional beings that wish to stay on this one-dimensional road. A region on this road is said to be habitable if and only if it can get at least "k" signals from "k" different transmitters. Given the road and the transmitters and their signal radius and the value of "k", we need to find out all the regions that can be considered habitable.
Assumptions:
1) There is no concept of signal strength in our world. A signal will have full strength up to its radius "r" and beyond that, it will have zero strength.
2) We have considered, without the loss of generality in the algorithm, the coordinates and the ranges of the transmitters to be purely integers. There will be no change in the algorithm in case we take them as floats, or doubles.
3) There will not be more than one transmitter on a point. If that is the case, the transmitter whose signal can travel farthest will be considered.
Brute Force Approach:
The brute force approach would be to keep a set of each n transmitters and the regions to which each can provide signals. In this case, the required answer would be the intersection of any k sets of the n total sets just formed. The time complexity of this approach is O(nCk)
O(nLogn) Approach:
Algorithm Axiom: Solutions become much more efficient when the data is arranged in some particular order.
Imagine us being a part of the one-dimensional world mentioned above. We have an infinite road in front of us. Beginning our search from somewhere in the middle is difficult because we do not know about the transmitters to our left and right. So we can make no sound decision about which direction to start looking for habitable regions in, If however, say, someone tells us that the region we are currently in is the left-most end of the range of the left-most transmitter, we know for a fact that there can be no habitable regions to our left. This makes our next step clear - to go to the right in the search of habitable regions.
Coming back to our algorithm, if we sort the input transmitters in the non-decreasing order of their coordinates. If we start from the first transmitter, we will traverse the list of transmitters in one direction, along with keeping a track of whether the region we are currently in habitable or not. The only question that now remains is how to keep track of habitable regions. It is impossible to keep a count of all the signals available at every point on the road, as there are infinite points.
Point to be noted: All habitable regions will begin from a transmitter's left-most signal limit and end at (possible another) transmitter's right-most signal limit.
Our task is simplified if we take into consideration the above point. To identify a region, we only require its start and end points, and these points will only be the let and right limits of some transmitters. Hence, instead of analyzing all points, we analyze only those points where a transmitters range ends or begins. There are maximum "2n" such points for "n" transmitters.
Algorithm NTransmitters (transmittersList, Integer k):
1) sortAscending(transmittersList) on their coordinates;
2) beginsEnds := list of all the left-limits and right-limits of all the transmitter signals;
3) Traverse only begin and end lists from beginsEnds[0]; //beginsEnds[0] will hold the left-most end of the road where a signal is being received. The traversal will not include all the points, just the "begin" and "end" points of the list.
4) If beginsEnds[i] = "begin":
count := count+1
Else:
count := count-1
5) If count >= k :
output region as habitable;
Following is a well-commented Java code for the above algorithm. The only other noteworthy thing is that we have maintained a TreeMap instead of the usual HashMap. TreeMap extends HashMap, but sorts the data on their key value. This is needed so that we start from the left-most begin point. (it will be the one with the least coordinate value). Also, we have maintained 2 separate lists of begin and end to optimize memory requirements, As always, anyone is free to ask any doubts or share any comments in the comments section.
There is a one-dimensional road of infinite length, which extends from -∞ to
There is a group of one-dimensional beings that wish to stay on this one-dimensional road. A region on this road is said to be habitable if and only if it can get at least "k" signals from "k" different transmitters. Given the road and the transmitters and their signal radius and the value of "k", we need to find out all the regions that can be considered habitable.
Assumptions:
1) There is no concept of signal strength in our world. A signal will have full strength up to its radius "r" and beyond that, it will have zero strength.
2) We have considered, without the loss of generality in the algorithm, the coordinates and the ranges of the transmitters to be purely integers. There will be no change in the algorithm in case we take them as floats, or doubles.
3) There will not be more than one transmitter on a point. If that is the case, the transmitter whose signal can travel farthest will be considered.
Brute Force Approach:
The brute force approach would be to keep a set of each n transmitters and the regions to which each can provide signals. In this case, the required answer would be the intersection of any k sets of the n total sets just formed. The time complexity of this approach is O(nCk)
O(nLogn) Approach:
Algorithm Axiom: Solutions become much more efficient when the data is arranged in some particular order.
Imagine us being a part of the one-dimensional world mentioned above. We have an infinite road in front of us. Beginning our search from somewhere in the middle is difficult because we do not know about the transmitters to our left and right. So we can make no sound decision about which direction to start looking for habitable regions in, If however, say, someone tells us that the region we are currently in is the left-most end of the range of the left-most transmitter, we know for a fact that there can be no habitable regions to our left. This makes our next step clear - to go to the right in the search of habitable regions.
Coming back to our algorithm, if we sort the input transmitters in the non-decreasing order of their coordinates. If we start from the first transmitter, we will traverse the list of transmitters in one direction, along with keeping a track of whether the region we are currently in habitable or not. The only question that now remains is how to keep track of habitable regions. It is impossible to keep a count of all the signals available at every point on the road, as there are infinite points.
Point to be noted: All habitable regions will begin from a transmitter's left-most signal limit and end at (possible another) transmitter's right-most signal limit.
Our task is simplified if we take into consideration the above point. To identify a region, we only require its start and end points, and these points will only be the let and right limits of some transmitters. Hence, instead of analyzing all points, we analyze only those points where a transmitters range ends or begins. There are maximum "2n" such points for "n" transmitters.
Algorithm NTransmitters (transmittersList, Integer k):
1) sortAscending(transmittersList) on their coordinates;
2) beginsEnds := list of all the left-limits and right-limits of all the transmitter signals;
3) Traverse only begin and end lists from beginsEnds[0]; //beginsEnds[0] will hold the left-most end of the road where a signal is being received. The traversal will not include all the points, just the "begin" and "end" points of the list.
4) If beginsEnds[i] = "begin":
count := count+1
Else:
count := count-1
5) If count >= k :
output region as habitable;
Following is a well-commented Java code for the above algorithm. The only other noteworthy thing is that we have maintained a TreeMap instead of the usual HashMap. TreeMap extends HashMap, but sorts the data on their key value. This is needed so that we start from the left-most begin point. (it will be the one with the least coordinate value). Also, we have maintained 2 separate lists of begin and end to optimize memory requirements, As always, anyone is free to ask any doubts or share any comments in the comments section.
