1385. Find the Distance Value Between Two Arrays

-
Quesition Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays. The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d. Example 1: Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2 Output: 2 Explanation: For arr1[0]=4 we have: |4-10|=6 > d=2 |4-9|=5 > d=2 |4-1|=3 > d=2 |4-8|=4 > d=2 For arr1[1]=5 we have: |5-10|=5 > d=2 |5-9|=4 > d=2 |5-1|=4 > d=2 |5-8|=3 > d=2 For arr1[2]=8 we have: |8-10|=2 <= d=2 |8-9|=1 <= d=2 |8-1|=7 > d=2 |8-8|=0 <= d=2 Example 2: Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3 Output: 2 Example 3: Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6 Output: 1 Suoltion class Solution { public: int findTheDistanceValue(vector& arr1, vector& arr2, int d) { int value = 0; int n = arr1.size(), m = arr2.size(); for(int i = 0; i < n;i++){ int j = 0; for(; j < m;j++){ if(abs(arr1[i] - arr2[j]) <= d) break; } if(j == m) value++; } return value; } };

Comments

Post a Comment

Popular posts from this blog

67. Add Binary

-
Add Binary   (C++) class Solution { public : string addBinary ( string a, string b ) { string ans= "" ; int carry= 0 ; while (a.size()||b.size()) { int x= 0 ,y= 0 ; if (a.size()){ x=a.back()- '0' ; a.pop_back(); } if (b.size()){ x+=b.back()- '0' ; b.pop_back(); } x+=carry; if (x== 1 ){ans+= '1' ;carry= 0 ;} if (x== 0 )ans+= '0' ; if (x== 2 ){ans+= '0' ;carry= 1 ;} if (x== 3 ){ans+= '1' ;carry= 1 ;} } if (carry)ans+= '1' ; reverse(ans.begin(),ans.end()); return ans; } };
Read more

160. Intersection of Two Linked Lists

-
Question Given the heads of two separately linked lists headA and headB , return the node at which the two lists intersect. If the two linked lists do not intersect at all, return null. For example, the following two linked lists begin to intersect at node c1: Test cases are generated in such a way that there are no loops anywhere in the entire linked structure. Note that linked lists must retain their original structure when the function returns. Own referee: The inputs for the referee are as follows (your program will not receive these inputs): intersectVal - The value of the node where the intersection occurs. This is 0 if there is no intersecting node. listA - The first linked list. listB - Second linked list. skipA - The number of nodes to skip forward in list A (starting from the beginning) to reach the intersected node. skipB - The number of nodes to skip forward in list B (starting from the beginning) to reach the intersecting node. The arbiter then...
Read more

1769. Minimum Number of Operations to Move All Balls to Each Box

-
Question You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the ith box is empty, and '1' if it contains one ball. In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some boxes. Return an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the ith box. Each answer[i] is calculated considering the initial state of the boxes. Example 1: Input: boxes = "110" Output: [1,1,3] Explanation: The answer for each box is as follows: 1) First box: you will have to move one ball from the second box to the first box in one operation. 2) Second box: you will have to move one ball from the first box to the second box in one operation. 3) Third box: you will have to move one ball from the first box to the third box in two operations, and move one ball fr...
Read more