Given an array arr, replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1.
After doing so, return the array.
Example 1:
Input: arr = [17,18,5,4,6,1]
Output: [18,6,6,6,1,-1]
Explanation:
- index 0 --> the greatest element to the right of index 0 is index 1 (18).
- index 1 --> the greatest element to the right of index 1 is index 4 (6).
- index 2 --> the greatest element to the right of index 2 is index 4 (6).
- index 3 --> the greatest element to the right of index 3 is index 4 (6).
- index 4 --> the greatest element to the right of index 4 is index 5 (1).
- index 5 --> there are no elements to the right of index 5, so we put -1.
Example 2:
Input: arr = [400]
Output: [-1]
Explanation: There are no elements to the right of index 0.
Constraints:
-
1 <= arr.length <= 104 -
1 <= arr[i] <= 105
SOLUTION:
class Solution:
# def makeSeg(self, arr, i, j):
# seg = self.seg
# if (i, j) in seg:
# return seg[(i, j)]
# if i == j:
# seg[(i, j)] = arr[i]
# return arr[i]
# mid = (i + j) // 2
# curr = max(self.makeSeg(arr, i, mid), self.makeSeg(arr, mid + 1, j))
# seg[(i, j)] = curr
# return curr
# def getMax(self, arr, i, j, ni, nj):
# seg = self.seg
# if ni >= i and nj <= j:
# return seg[(ni, nj)]
# if (ni < i and nj < i) or (ni > j and nj > j):
# return float('-inf')
# mid = (ni + nj) // 2
# return max(self.getMax(arr, i, j, ni, mid), self.getMax(arr, i, j, mid + 1, nj))
# def replaceElements(self, arr: List[int]) -> List[int]:
# n = len(arr)
# self.seg = {}
# self.makeSeg(arr, 0, n - 1)
# for i in range(n):
# if i < n - 1:
# arr[i] = self.getMax(arr, i + 1, n, 0, n - 1)
# else:
# arr[i] = -1
# return arr
def replaceElements(self, arr: List[int]) -> List[int]:
n = len(arr)
op = [-1] * n
currMax = float('-inf')
for i in range(n - 1, 0, -1):
currMax = max(currMax, arr[i])
op[i - 1] = currMax
return op
Top comments (0)