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  • + 0 comments

    def beautifulDays(i, j, k): count = 0 for r in range(i,j+1): str1 = str(r) rev_str = str1[::-1] rev_num = int(rev_str) diff = abs(r-rev_num) num = diff/k num1 = diff//k if num == num1: count += 1 return count

  • + 0 comments

    my very easy cpp solution

    int beautifulDays(int i, int j, int k) {

    int rem,count=0,temp;
    
    for(int m=i;m <=j ; m++){///for looping
        temp=m;
        int rev=0;
        while(temp > 0){
        }
    
        if(abs(m - rev) % k ==0){  \\for checking it is equallly divisible or not
            count++;
    
        }
    }
    return count;
    

    }

  • + 0 comments

    my very easy cpp solution

    int beautifulDays(int i, int j, int k) {

    int rem,count=0,temp;
    
    for(int m=i;m <=j ; m++){///for looping
        temp=m;
        int rev=0;
        while(temp > 0){
             rem = temp % 10;
             rev = (rev * 10 ) + rem;
             temp = temp/10;  //for reversing number
        }
    
        if(abs(m - rev) % k ==0){  \\for checking it is equallly divisible or not
            count++;
    
        }
    }
    return count;
    

    }

  • + 0 comments

    Here is my solution in JavaScript:

    ` let count = 0

    for(let n = i; n <= j; n++){

    const reversed = parseInt(n.toString().split('').reverse().join(''), 10);

    const diff = Math.abs(n - reversed);

    if(diff % k === 0){
    
        count++
    }
    

    }

    return count; `

  • + 0 comments

    Here is problem solution in Python, Java, C++, C and javascript - https://programmingoneonone.com/hackerrank-beautiful-days-at-the-movies-problem-solution.html