Welcome to Subscribe On Youtube

3751. Total Waviness of Numbers in Range I

Description

You are given two integers num1 and num2 representing an inclusive range [num1, num2].

The waviness of a number is defined as the total count of its peaks and valleys:

• A digit is a peak if it is strictly greater than both of its immediate neighbors.
• A digit is a valley if it is strictly less than both of its immediate neighbors.
• The first and last digits of a number cannot be peaks or valleys.
• Any number with fewer than 3 digits has a waviness of 0.

Return the total sum of waviness for all numbers in the range [num1, num2].

 

Example 1:

Input: num1 = 120, num2 = 130

Output: 3

Explanation:

In the range [120, 130]:

• 120: middle digit 2 is a peak, waviness = 1.
• 121: middle digit 2 is a peak, waviness = 1.
• 130: middle digit 3 is a peak, waviness = 1.
• All other numbers in the range have a waviness of 0.

Thus, total waviness is 1 + 1 + 1 = 3.

Example 2:

Input: num1 = 198, num2 = 202

Output: 3

Explanation:

In the range [198, 202]:

• 198: middle digit 9 is a peak, waviness = 1.
• 201: middle digit 0 is a valley, waviness = 1.
• 202: middle digit 0 is a valley, waviness = 1.
• All other numbers in the range have a waviness of 0.

Thus, total waviness is 1 + 1 + 1 = 3.

Example 3:

Input: num1 = 4848, num2 = 4848

Output: 2

Explanation:

Number 4848: the second digit 8 is a peak, and the third digit 4 is a valley, giving a waviness of 2.

 

Constraints:

• 1 <= num1 <= num2 <= 105

Solutions

Solution 1: Simulation

We define a helper function $f(x)$ to calculate the waviness value of integer $x$. In this function, we store each digit of integer $x$ in an array $\textit{nums}$. If the number has fewer than 3 digits, the waviness value is 0. Otherwise, we iterate through each non-leading and non-trailing digit in the array $\textit{nums}$, determine whether it is a peak or valley, and count the waviness value.

Then, we iterate through each integer $x$ in the range $[\textit{num1}, \textit{num2}]$ and accumulate its waviness value $f(x)$ to obtain the final result.

The time complexity is $O((\textit{num2} - \textit{num1} + 1) \cdot \log \textit{num2})$ and the space complexity is $O(\log \textit{num2})$.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int totalWaviness(int num1, int num2) {
          int ans = 0;
          for (int x = num1; x <= num2; x++) {
              ans += f(x);
          }
          return ans;
      }
  
      private int f(int x) {
          int[] nums = new int[20];
          int m = 0;
          while (x > 0) {
              nums[m++] = x % 10;
              x /= 10;
          }
          if (m < 3) {
              return 0;
          }
          int s = 0;
          for (int i = 1; i < m - 1; i++) {
              if ((nums[i] > nums[i - 1] && nums[i] > nums[i + 1])
                  || (nums[i] < nums[i - 1] && nums[i] < nums[i + 1])) {
                  s++;
              }
          }
          return s;
      }
  }
  
  

• class Solution {
  public:
      int totalWaviness(int num1, int num2) {
          int ans = 0;
          for (int x = num1; x <= num2; x++) {
              ans += f(x);
          }
          return ans;
      }
  
      int f(int x) {
          int nums[20], m = 0;
          while (x > 0) {
              nums[m++] = x % 10;
              x /= 10;
          }
          if (m < 3) {
              return 0;
          }
          int s = 0;
          for (int i = 1; i < m - 1; i++) {
              if ((nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) || (nums[i] < nums[i - 1] && nums[i] < nums[i + 1])) {
                  s++;
              }
          }
          return s;
      }
  };
  
  

• class Solution:
      def totalWaviness(self, num1: int, num2: int) -> int:
          def f(x: int) -> int:
              nums = []
              while x:
                  nums.append(x % 10)
                  x //= 10
              m = len(nums)
              if m < 3:
                  return 0
              s = 0
              for i in range(1, m - 1):
                  if nums[i] > nums[i - 1] and nums[i] > nums[i + 1]:
                      s += 1
                  elif nums[i] < nums[i - 1] and nums[i] < nums[i + 1]:
                      s += 1
              return s
  
          return sum(f(x) for x in range(num1, num2 + 1))
  
  

• func totalWaviness(num1 int, num2 int) (ans int) {
  	for x := num1; x <= num2; x++ {
  		ans += f(x)
  	}
  	return
  }
  
  func f(x int) int {
  	nums := make([]int, 0, 20)
  	for x > 0 {
  		nums = append(nums, x%10)
  		x /= 10
  	}
  	m := len(nums)
  	if m < 3 {
  		return 0
  	}
  	s := 0
  	for i := 1; i < m-1; i++ {
  		if (nums[i] > nums[i-1] && nums[i] > nums[i+1]) ||
  			(nums[i] < nums[i-1] && nums[i] < nums[i+1]) {
  			s++
  		}
  	}
  	return s
  }
  
  

• function totalWaviness(num1: number, num2: number): number {
      let ans = 0;
      for (let x = num1; x <= num2; x++) {
          ans += f(x);
      }
      return ans;
  }
  
  function f(x: number): number {
      const nums: number[] = [];
      while (x > 0) {
          nums.push(x % 10);
          x = Math.floor(x / 10);
      }
      const m = nums.length;
      if (m < 3) return 0;
  
      let s = 0;
      for (let i = 1; i < m - 1; i++) {
          if (
              (nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) ||
              (nums[i] < nums[i - 1] && nums[i] < nums[i + 1])
          ) {
              s++;
          }
      }
      return s;
  }
  
  

All Problems

All Solutions