package _interview_question
/**
* Problem: Find the most frequent element in all intervals
Given an unsorted list of start and end time ( a range basically), find any number within all the ranges that occurs in maximum number of intervals.
Example: [[1,4],[3,5],[4,6]].
Answer should be 4 because it occurs in all the interval ranges.
* */
class Solution6 {
/*
* solution 1:brute force, Time complexity:O(n*n), Space complexity:O(n)
* */
fun findMostFrequent(intervals: Array<IntArray>): Int {
val map = HashMap<Int, Int>()
for (interval in intervals) {
for (item in interval) {
map.put(item, map.getOrDefault(item, 0) + 1)
}
}
var maxValue = Int.MIN_VALUE
for (item in map) {
maxValue = Math.max(maxValue, item.value)
}
var result = 0
for (item in map) {
if (maxValue == item.value) {
result = item.key
}
}
println(map)
return result
}
/*
* solution 2:Time complexity:O(n), Space complexity:O(n);
1.Initialize a freq array with 0, let the size of the array be 10^6 as this is the maximum possible.
2.Increase the freq[l] by 1, for every starting index of the given range.
3.Decrease the freq[r+1] by 1 for every ending index of the given range.
4.Iterate from the minimum Left to the maximum Right and add the frequencies by freq[i] += freq[i-1].
5.The index with the maximum value of freq[i] will be the answer.
* */
fun findMostFrequent2(intervals: Array<IntArray>): Int {
val freq = IntArray(1000000)
var first = 0
var last = 0
/*
* eg: { {1, 6}, {2, 3}, {2, 5}, {3, 8} }
{
0: 0
1: 1
2: 2
3: 1
4: -1
7: -1
6: -1
9: -1
}
change into:
1 -> 0 + 1
2 -> 1 + 2
3 -> 4
4 -> 3
5 -> 3
6 -> 2
7 -> 1
8 -> 1
* */
for (i in intervals.indices) {
val left = intervals[i][0]
val right = intervals[i][1]
freq[left]++
freq[right + 1]--
first = Math.min(left, first)
last = Math.max(right, last)
}
var maxFrequent = 0
var num = 0
//check for most frequent element
for (i in first + 1..last) {
//increase the frequency
freq[i] = freq[i - 1] + freq[i]
if (freq[i] > maxFrequent) {
maxFrequent = freq[i]
num = i
}
}
return num
}
}
