Smallest Range II

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

    Example 1: 

    Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

    Example 2: 

    Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

    Example 3: 

    Input: A = [1,3,6], K = 3
Output: 3
Explanation: B = [4,6,3]

    Note:

    1. 1 <= A.length <= 10000
    2. 0 <= A[i] <= 10000
    3. 0 <= K <= 10000
    Solution:

    class Solution {
        public int smallestRangeII(int[] A, int K) {
            int i, min=0, max=0, min1, min2, max1, max2, length = A.length, diff=0;
            Arrays.sort(A);
            diff = A[length-1]-A[0];
            int result = Integer.MAX_VALUE;
            
                min1 = A[0]+K;
                max1 = A[length-1]-K;
            
            for(i=1;i<length;i++){
                min2 = A[i]-K;
                max2 = A[i-1]+K;
                min = Math.min(min1, min2);
                max = Math.max(max1, max2);
                result = Math.min(result, max-min);
            }
            return Math.min(diff, result);
        }
    }


    1) Initially sort the array, so that we can find one of the best difference between min value and max value in the given array.(i.e., diff variable in above code)
    2) min1 is a variable which holds smallest value from a given array and also along with the addition of K.
    3) max1 is a variable which holds largest value from a given array and also along with the subtraction of K.
    4) min2 and max2 variables change accordingly based on iteration. min2 checks min value for current i'th position value. max2 checks max value for previous i'th position value. So, for each value in a array we check for both +K and -K value.
    5) Find min value from min1 and min2
    6) Find max value from max1 and max2.
    7) Find min from result and max,min difference and store it in result.
    8) Finally, return min among result and diff.

    Related Posts:


    Like us? Please do share with your friends ..!!

    Follow us to receive updates instantly.

    If you have any feedback/suggestions, leave it in a comment section or contact us through our contact page which will be helpful for us to improve.


    The intention behind this blog is to help all my fellow programmers. Also to provide solutions for various coding platform problems with solution and explanation. I will add many tutorials in future. I am not a professional blogger but whenever time permits I will post. I hope you’ll enjoy the content and find it useful! Check out my Writer blog for a more valuable quotes. For any queries or suggestions, please feel free to reach out to me. Happy coding!

    Comments

    Post a Comment

    Popular posts from this blog

    First Unique Character in a String

    Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase letters. class Solution {     public int firstUniqChar(String s) {         int length = s.length();         String str;         for(int i=0;i<length;i++){             if(i==length-1){                 str = s.substring(0, i);             }             else{                 str = s.substring(0,i)+s.substring(i+1);             }             if(!str.contains(s.charAt(i)+""))                 return i;         }         return -1;     } } Try it on Leetcode Here, we are going to use substring method. 1) If character is at end, then take substring from 0 till before last character of string. 2) Else make a temporary string, by removing exactly that particular character using index. [e.g..
    Read more

    Balanced Binary Tree

    Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of  every  node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true   Constraints: The number of nodes in the tree is in the range  [0, 5000] . -10 4  <= Node.val <= 10 4 Solution: /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }
    Read more

    Majority Element

    Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int repeatLen
    Read more