Given n ranges of the form L and R, the task is to find the maximum occurred integer in all the ranges. If more than one such integer exits, print the smallest one.
0 <= Li, Ri < 1000000.
Input : L1 = 1 R1 = 15 L2 = 4 R2 = 8 L3 = 3 R3 = 5 L3 = 1 R3 = 4 Output : 4 Input : L1 = 1 R1 = 15 L2 = 5 R2 = 8 L3 = 9 R3 = 12 L4 = 13 R4 = 20 L5 = 21 R5 = 30 Output : 5 Numbers having maximum occurrence i.e 2 are 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. The smallest number among all are 5.
A simple solution is to use hash table to store counts of all numbers. We traverse every interval from Li to Ri and increment count of all numbers present in every interval. Finally we traverse the hash table to find the number with maximum count. Time complexity of this solution is O(n * MAX_INTERVAL) where MAX_INTERVAL is maximum number of elements in an interval.
An efficient solution requires linear time. We create an array arr of size 1000000 (limit given on maximum value of an interval). The idea is to add +1 to each Li index and -1 to corresponding Ri index in arr. After this, find the prefix sum of the array. Adding +1 at Li shows the starting point of ith Range and adding -1 shows the ending point of ith range. Finally we return the array index that has maximum prefix sum
Algorithm to solve the problem:
- Initialize an array arr to 0.
- For each range i, add +1 at Li index and -1 at Ri of the array.
- Find the prefix sum of the array and find the smallest index having maximum prefix sum.
Below is the implementation of this approach:
Time Complexity: O(n + MAX)
Exercise: Try for 0 <= Li, Ri <= 1000000000. (Hint: Use stl map).
Related Article : Maximum value in an array after m range increment operations
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