441. Arranging Coins

-

 class Solution {

public:
    int arrangeCoins(int n) {
    int i = 0 ;
    for( ; n-i>=0 ; i++ ) n=n-i;
    return i-1;       
    }
};

Comments

Post a Comment

Popular posts from this blog

2161. Partition Array According to Given Pivot

-
Question You are given a 0-indexed integer array nums and an integer pivot. Rearrange nums such that the following conditions are satisfied: Every element less than pivot appears before every element greater than pivot. Every element equal to pivot appears in between the elements less than and greater than pivot. The relative order of the elements less than pivot and the elements greater than pivot is maintained. More formally, consider every pi, pj where pi is the new position of the ith element and pj is the new position of the jth element. For elements less than pivot, if i pivot and nums[j] > pivot, then pi pivotArray(vector & nums, int pivot) { vector s; vector l; int d=0; for(int i=0;i
Read more

2000. Reverse Prefix of Word

-
Question Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing. For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd". Return the resulting string. Example 1: Input: word = "abcdefd", ch = "d" Output: "dcbaefd" Explanation: The first occurrence of "d" is at index 3. Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd". Example 2: Input: word = "xyxzxe", ch = "z" Output: "zxyxxe" Explanation: The first and only occurrence of "z" is at index 3. Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe". Solution class Solution ...
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