Replace Elements with Greatest Element on Right Side
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 <= 1041 <= arr[i] <= 105
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
Comments
Post a Comment