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This approach involves sorting the array first. After sorting, we iterate over the array from the first element onwards, setting each element to be the minimum between its current value and the value of the previous element plus 1. This ensures the conditions are met, i.e., the first element is 1 and the difference between adjacent elements is <= 1.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) as we modify the array in place.
1def maxElementAfterDecreasingAndRearranging(arr):
2 arr.sort()
3 arr[0] = 1
4 for i in range(1, len(arr)):
5 arr[i] = min(arr[i], arr[i-1] + 1)
6 return arr[-1]
7
8arr = [100, 1, 1000]
9print(maxElementAfterDecreasingAndRearranging(arr))Python provides a straightforward implementation using the sort method, which organizes the array in non-decreasing order. From there, we iterate and use min to adjust values to fit the requirements.
This approach leverages the counting sort technique which avoids sorting the entire array directly. Instead, it uses a frequency array to count occurrences of each element. Then, reconstruct the final array by checking counts and adjusting values accordingly to ensure the maximum element is achieved while meeting the constraints.
Time Complexity: O(n), due to the pass over the array to count elements.
Space Complexity: O(n), primarily due to the count array.
#include <vector>
using namespace std;
int maxElementAfterDecreasingAndRearranging(vector<int>& arr) {
int n = arr.size();
vector<int> count(n + 1, 0);
for (int num : arr)
count[min(num, n)]++;
int maxElem = 0;
for (int i = 1; i <= n; i++) {
maxElem += count[i];
if (maxElem < i) maxElem = i;
}
return maxElem;
}
int main() {
vector<int> arr = {100, 1, 1000};
cout << maxElementAfterDecreasingAndRearranging(arr);
return 0;
}C++ implementation utilizes a vector as a counting array. Each element in arr updates this vector, allowing construction of an adjusted arr satisfying the conditions without explicit initial sorting.