962. Maximum Width Driveway – DEV Community

962. Maximum width of ramp

Difficulty: Medium

Topics: Array, Stack, Monotonic Stack

A driveway in an integer array nums is a few (i, j) for what i and nums(i) . The width of such a ramp is j - i.

Given an integer array numsyield the maximum width of a driveway in nums. If there is none driveway in numsyield 0.

Example 1:

• Input: num = (6,0,8,2,1,5)
• Exit: 4
• Explanation: The maximum width slope is reached at (i, j) = (1, 5): nums(1) = 0 and nums(5) = 5.

Example 2:

• Input: num = (9,8,1,0,1,9,4,0,4,1)
• Exit: 7
• Explanation: The maximum width slope is reached at (i, j) = (2, 9): nums(2) = 1 and nums(9) = 1.

Limits:

Solution:

We can use the concept of a monotonous pile. Here is the solution and explanation:

Approximation:

1. Monotonous decreasing stack: We create a stack that keeps track of the indices of elements in such a way nums(stack(i)) is in descending order. This will allow us to find pairs later (i, j) Where nums(i) .
2. Traverse from the end: After creating the stack, we traverse the array from the end (j by n-1 Unpleasant 0) to try to find the furthest i for each j Where nums(i) .
3. Update the maximum width: Whenever nums(i) holds for the current top of the stack, calculate the ramp width j - i and update the maximum width if it is larger.

Let’s implement this solution in PHP: 962. Maximum width of ramp

/**
 * @param Integer() $nums
 * @return Integer
 */
function maxWidthRamp($nums) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example 1
$nums = (6, 0, 8, 2, 1, 5);
echo maxWidthRamp($nums);  // Output: 4

// Example 2
$nums = (9, 8, 1, 0, 1, 9, 4, 0, 4, 1);
echo maxWidthRamp($nums);  // Output: 7
?>

Explanation:

1. Create a decreasing pile:

   • Iterate through the array and add indices to the stack.
   • Add an index only if it matches a value less than or equal to the value of the last index in the stack. This ensures that the values ​​in the stack are in descending order.

2. Traverse from the end:

   • As we move backward through the array, for each jpop indexes i of the pile as long as nums(i) .
   • Calculate the width j - i and update maxWidth.

3. Why this works:

   • By maintaining a decreasing stack of indices, we ensure that when we have a j with a larger value it can give us a larger width j - i when i is picked from the pile.

4. Time complexity:

   • Building the stack takes time O(n) time since each index was pushed once.
   • The passage from the end and the popping indices also last O(n) as each index is displayed at most once.
   • In general, the solution comes in O(n) time, which is efficient for input sizes up to 5 * 10^4.

Exit:

• For nums = (6, 0, 8, 2, 1, 5)is the output 4corresponding to the driveway (1, 5).
• For nums = (9, 8, 1, 0, 1, 9, 4, 0, 4, 1)is the output 7corresponding to the driveway (2, 9).

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