| comments | difficulty | edit_url | rating | source | tags | |||
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true |
Hard |
1827 |
Weekly Contest 142 Q3 |
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(This problem is an interactive problem.)
You may recall that an array arr is a mountain array if and only if:
arr.length >= 3- There exists some
iwith0 < i < arr.length - 1such that:arr[0] < arr[1] < ... < arr[i - 1] < arr[i]arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index does not exist, return -1.
You cannot access the mountain array directly. You may only access the array using a MountainArray interface:
MountainArray.get(k)returns the element of the array at indexk(0-indexed).MountainArray.length()returns the length of the array.
Submissions making more than 100 calls to MountainArray.get will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.
Example 1:
Input: array = [1,2,3,4,5,3,1], target = 3 Output: 2 Explanation: 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.
Example 2:
Input: array = [0,1,2,4,2,1], target = 3
Output: -1
Explanation: 3 does not exist in the array, so we return -1.
Constraints:
3 <= mountain_arr.length() <= 1040 <= target <= 1090 <= mountain_arr.get(index) <= 109
# """
# This is MountainArray's API interface.
# You should not implement it, or speculate about its implementation
# """
# class MountainArray:
# def get(self, index: int) -> int:
# def length(self) -> int:
class Solution:
def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
def search(l: int, r: int, k: int) -> int:
while l < r:
mid = (l + r) >> 1
if k * mountain_arr.get(mid) >= k * target:
r = mid
else:
l = mid + 1
return -1 if mountain_arr.get(l) != target else l
n = mountain_arr.length()
l, r = 0, n - 1
while l < r:
mid = (l + r) >> 1
if mountain_arr.get(mid) > mountain_arr.get(mid + 1):
r = mid
else:
l = mid + 1
ans = search(0, l, 1)
return search(l + 1, n - 1, -1) if ans == -1 else ans/**
* // This is MountainArray's API interface.
* // You should not implement it, or speculate about its implementation
* interface MountainArray {
* public int get(int index) {}
* public int length() {}
* }
*/
class Solution {
private MountainArray mountainArr;
private int target;
public int findInMountainArray(int target, MountainArray mountainArr) {
int n = mountainArr.length();
int l = 0, r = n - 1;
while (l < r) {
int mid = (l + r) >>> 1;
if (mountainArr.get(mid) > mountainArr.get(mid + 1)) {
r = mid;
} else {
l = mid + 1;
}
}
this.mountainArr = mountainArr;
this.target = target;
int ans = search(0, l, 1);
return ans == -1 ? search(l + 1, n - 1, -1) : ans;
}
private int search(int l, int r, int k) {
while (l < r) {
int mid = (l + r) >>> 1;
if (k * mountainArr.get(mid) >= k * target) {
r = mid;
} else {
l = mid + 1;
}
}
return mountainArr.get(l) == target ? l : -1;
}
}/**
* // This is the MountainArray's API interface.
* // You should not implement it, or speculate about its implementation
* class MountainArray {
* public:
* int get(int index);
* int length();
* };
*/
class Solution {
public:
int findInMountainArray(int target, MountainArray& mountainArr) {
int n = mountainArr.length();
int l = 0, r = n - 1;
while (l < r) {
int mid = (l + r) >> 1;
if (mountainArr.get(mid) > mountainArr.get(mid + 1)) {
r = mid;
} else {
l = mid + 1;
}
}
auto search = [&](int l, int r, int k) -> int {
while (l < r) {
int mid = (l + r) >> 1;
if (k * mountainArr.get(mid) >= k * target) {
r = mid;
} else {
l = mid + 1;
}
}
return mountainArr.get(l) == target ? l : -1;
};
int ans = search(0, l, 1);
return ans == -1 ? search(l + 1, n - 1, -1) : ans;
}
};/**
* // This is the MountainArray's API interface.
* // You should not implement it, or speculate about its implementation
* type MountainArray struct {
* }
*
* func (this *MountainArray) get(index int) int {}
* func (this *MountainArray) length() int {}
*/
func findInMountainArray(target int, mountainArr *MountainArray) int {
n := mountainArr.length()
l, r := 0, n-1
for l < r {
mid := (l + r) >> 1
if mountainArr.get(mid) > mountainArr.get(mid+1) {
r = mid
} else {
l = mid + 1
}
}
search := func(l, r, k int) int {
for l < r {
mid := (l + r) >> 1
if k*mountainArr.get(mid) >= k*target {
r = mid
} else {
l = mid + 1
}
}
if mountainArr.get(l) == target {
return l
}
return -1
}
ans := search(0, l, 1)
if ans == -1 {
return search(l+1, n-1, -1)
}
return ans
}/**
* // This is the MountainArray's API interface.
* // You should not implement it, or speculate about its implementation
* class Master {
* get(index: number): number {}
*
* length(): number {}
* }
*/
function findInMountainArray(target: number, mountainArr: MountainArray) {
const n = mountainArr.length();
let l = 0;
let r = n - 1;
while (l < r) {
const mid = (l + r) >> 1;
if (mountainArr.get(mid) > mountainArr.get(mid + 1)) {
r = mid;
} else {
l = mid + 1;
}
}
const search = (l: number, r: number, k: number): number => {
while (l < r) {
const mid = (l + r) >> 1;
if (k * mountainArr.get(mid) >= k * target) {
r = mid;
} else {
l = mid + 1;
}
}
return mountainArr.get(l) === target ? l : -1;
};
const ans = search(0, l, 1);
return ans === -1 ? search(l + 1, n - 1, -1) : ans;
}impl Solution {
#[allow(dead_code)]
pub fn find_in_mountain_array(target: i32, mountain_arr: &MountainArray) -> i32 {
let n = mountain_arr.length();
// First find the maximum element in the array
let mut l = 0;
let mut r = n - 1;
while l < r {
let mid = (l + r) >> 1;
if mountain_arr.get(mid) > mountain_arr.get(mid + 1) {
r = mid;
} else {
l = mid + 1;
}
}
let left = Self::binary_search(mountain_arr, 0, l, 1, target);
if left == -1 {
Self::binary_search(mountain_arr, l, n - 1, -1, target)
} else {
left
}
}
#[allow(dead_code)]
fn binary_search(m: &MountainArray, mut l: i32, mut r: i32, k: i32, target: i32) -> i32 {
let n = m.length();
while l < r {
let mid = (l + r) >> 1;
if k * m.get(mid) >= k * target {
r = mid;
} else {
l = mid + 1;
}
}
if m.get(l) == target {
l
} else {
-1
}
}
}