Given an array A of integers, a ramp is a tuple (i, j) for which i < j and A[i] <= A[j]. The width of such a ramp is j - i.
Find the maximum width of a ramp in A. If one doesn't exist, return 0.
Example 1:
Input: [6,0,8,2,1,5]
Output: 4
Explanation:
The maximum width ramp is achieved at (i, j) = (1, 5): A[1] = 0 and A[5] = 5.
Example 2:
Input: [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation:
The maximum width ramp is achieved at (i, j) = (2, 9): A[2] = 1 and A[9] = 1.
Note:
2 <= A.length <= 50000
0 <= A[i] <= 50000
class Solution {
public int maxWidthRamp(int[] nums) {
int n = nums.length;
int[] rMax = new int[n];
rMax[n - 1] = nums[n - 1];
for (int i = n - 2; i >= 0; i--) {
rMax[i] = Math.max(rMax[i + 1], nums[i]);
}
// The trick is that left pointer iterates over original array and right pointer
// iterates over an array which stores maximum no. on the right for each index.
int left = 0, right = 0;
int ans = 0;
while (right < n) {
// If item at 'left' index > all the elements [right,n)
// then it cannot be used in our answer
while (left < right && nums[left] > rMax[right]) {
left++;
}
ans = Math.max(ans, right - left);
right++;
}
return ans;
}
}