LeetCode in Kotlin
1187. Make Array Strictly Increasing
Hard
Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing.
In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j].
If there is no way to make arr1 strictly increasing, return -1.
Example 1:
Input: arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
Output: 1
Explanation: Replace 5 with 2, then arr1 = [1, 2, 3, 6, 7].
Example 2:
Input: arr1 = [1,5,3,6,7], arr2 = [4,3,1]
Output: 2
Explanation: Replace 5 with 3 and then replace 3 with 4. arr1 = [1, 3, 4, 6, 7].
Example 3:
Input: arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
Output: -1
Explanation: You can’t make arr1 strictly increasing.
Constraints:
1 <= arr1.length, arr2.length <= 20000 <= arr1[i], arr2[i] <= 10^9
Solution
class Solution {
fun makeArrayIncreasing(arr1: IntArray, arr2: IntArray): Int {
arr2.sort()
var start = 0
for (i in arr2.indices) {
if (arr2[i] != arr2[start]) {
arr2[++start] = arr2[i]
}
}
val l2 = start + 1
val dp = IntArray(l2 + 2)
for (i in arr1.indices) {
var noChange = dp[dp.size - 1]
if (i > 0 && arr1[i - 1] >= arr1[i]) {
noChange = -1
}
for (j in dp.size - 2 downTo 1) {
if (arr2[j - 1] < arr1[i] && dp[j] != -1) {
noChange = if (noChange == -1) dp[j] else Math.min(noChange, dp[j])
}
if (dp[j - 1] != -1) {
dp[j] = 1 + dp[j - 1]
} else {
dp[j] = -1
}
if (i > 0 && arr1[i - 1] < arr2[j - 1] && dp[dp.size - 1] >= 0) {
dp[j] = if (dp[j] == -1) dp[dp.size - 1] + 1 else Math.min(dp[j], dp[dp.size - 1] + 1)
}
}
dp[0] = -1
dp[dp.size - 1] = noChange
}
var res = -1
for (num in dp) {
if (num != -1) {
res = if (res == -1) num else Math.min(res, num)
}
}
return res
}
}