🚀Day 17/180 1752. Check if Array Is Sorted and Rotated(Leetcode)

1752. Check if Array Is Sorted and Rotated

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class Solution {
    public boolean check(int[] nums) {
        int spike=0;
        for(int i=0;i<nums.length-1;i++){
            if(nums[i]>nums[i+1]) spike++;
        }
        if(nums[nums.length-1]>nums[0]) spike++;
        if(spike>1) return false;
        return true;
    }
}

DRY :

Dry run of the given solution with an example. We'll use the array [3, 4, 5, 1, 2] to demonstrate the dry run.

Array: [3, 4, 5, 1, 2]

Initial Setup

• spike = 0

Loop Through the Array

1. Iteration 1 (i = 0):

   • nums[0] = 3

   • nums[1] = 4

   • 3 <= 4 (No spike)

   • spike = 0

2. Iteration 2 (i = 1):

   • nums[1] = 4

   • nums[2] = 5

   • 4 <= 5 (No spike)

   • spike = 0

3. Iteration 3 (i = 2):

   • nums[2] = 5

   • nums[3] = 1

   • 5 > 1 (Spike detected)

   • spike = 1

4. Iteration 4 (i = 3):

   • nums[3] = 1

   • nums[4] = 2

   • 1 <= 2 (No spike)

   • spike = 1

Final Comparison

• Compare the last element with the first element:

  • nums[4] = 2

  • nums[0] = 3

  • 2 < 3 (No spike here)

  • spike = 1

Final Spike Check

• spike = 1, which is not greater than 1.

Result

• Since spike <= 1, the function returns true.

Conclusion

The array [3, 4, 5, 1, 2] is determined to be sorted and rotated.

Here's the step-by-step dry run in code format:

class Solution {
    public boolean check(int[] nums) {
        int spike = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] > nums[i + 1]) {
                spike++;
            }
        }
        if (nums[nums.length - 1] > nums[0]) {
            spike++;
        }
        return spike <= 1;
    }
}

// Dry run example
Solution solution = new Solution();
int[] nums = {3, 4, 5, 1, 2};
boolean result = solution.check(nums);
// result should be true

This dry run confirms that the solution correctly identifies the array [3, 4, 5, 1, 2] as sorted and rotated.