84 lines
2.2 KiB
Java
84 lines
2.2 KiB
Java
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/** Problem Description :
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* You are given a number N. Find the total count of set bits for all numbers from 1 to N
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* including 1 and N
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*/
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/**
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* Author : Suraj Kumar
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* Github : https://github.com/skmodi649
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*/
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/** ALGORITHM :
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* So, we will iterate till the number of bits in the number.
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* And we don’t have to iterate every single number in the range from 1 to n.
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* We will perform the following operations to get the desired result.
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* The quotient when n+1 is divided by 2 will return the number of times the 0,1 pattern has appeared at LSB.
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* However, the quotient when n+1 is divided by 4 will return the number of times the 0,0,1,1 pattern has
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* appeared at 2nd least significant bit and so on.
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*/
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/** TEST CASES :
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* Test Case 1 :
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* Input: N = 4
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* Output: 5
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*
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* Test Case 2 :
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* Input: N = 17
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* Output: 35
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*/
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import java.util.*;
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import java.lang.*;
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class Solution {
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//Function to return sum of count of set bits in the integers from 1 to n.
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public static int countSetBits(int n) {
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int count = 0 , i = 1 , val = 0;
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while ((n + 1) / (int) Math.pow(2, i) != 0) {
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int k = (n + 1) % (int) Math.pow(2, i);
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val = (n + 1) / (int) Math.pow(2, i);
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int c = (int) Math.pow(2, i - 1);
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count = count + c * val;
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if (k > ((int) Math.pow(2, i) / 2)) {
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count = count + (k - ((int) Math.pow(2, i) / 2));
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}
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i++;
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}
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int temp = (int) Math.pow(2, i);
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count += (n + 1 - temp / 2);
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return count;
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}
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public static void main(String[] args){
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Scanner sc = new Scanner(System.in);
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System.out.println("Enter the number :");
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int n = sc.nextInt();//input n
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Solution obj = new Solution();
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System.out.println("Total set bits : "+obj.countSetBits(n)); // calling countSetBits method
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System.out.println(); // changing the line
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}
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}
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/** Explanation for N = 4
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* For numbers from 1 to 4.
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* For 1: 0 0 1 = 1 set bits
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* For 2: 0 1 0 = 1 set bits
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* For 3: 0 1 1 = 2 set bits
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* For 4: 1 0 0 = 1 set bits
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* Therefore, the total set bits is 5.
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*/
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/** Time Complexity : O(log N)
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* Auxiliary Space Complexity : O(1)
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* Constraints : 1 ≤ N ≤ 10^8
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*/
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