{"id":762,"date":"2020-06-10T14:07:15","date_gmt":"2020-06-10T14:07:15","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=762"},"modified":"2022-04-06T10:32:25","modified_gmt":"2022-04-06T10:32:25","slug":"find-the-closest-palindrome","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/find-the-closest-palindrome\/","title":{"rendered":"Find the Closest Palindrome"},"content":{"rendered":"

\"\"<\/p>\n

Concepts Used<\/h3>\n
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Strings<\/p>\n<\/blockquote>\n

Difficulty Level<\/h3>\n
\n

Hard<\/p>\n<\/blockquote>\n

Problem Statement (Simplified):<\/h3>\n
\n

For given numbers we have to print the nearest anagram to it if two numbers are equidistant and anagram to the given number, print out the smallest. If the number is a palindrome itself, print its nearest palindrome.<\/p>\n<\/blockquote>\n

<\/strong><\/u><\/a><\/p>\n

Test Case<\/h4>\n
    Input:\n    3\n    6\n    127\n    121\n\n    Output:\n    5\n    131\n    111\n\n    Explanation:\n    Case-1:\n    5 is the nearest palindrome for 6.\n\n    Case-2:\n    127 has 121 and 131 as palindrome near to it. 131 is nearest to 127, so 131 is our answer.\n\n    Case-3:\n    121 has 111 and 131 as palindrome near to it. Both are equidistant to it. So we print the smallest one. Hence, 111 is our answer.<\/code><\/pre>\n
Solving Approach :<\/h3>\n
For checking out the nearest palindrome, we check different cases.<\/p>\n
    \n
  • Case -1:<\/code> If the number is a single digit, it answers will be one less to it. <\/li>\n
  • Case -2:<\/code> If the number contains only 9’s at every place, the answer will be the number+2 always.<\/li>\n
  • Case -3 :<\/code> If number is in form of 1000..{0,1}<\/code>, for example 10, 101, 100,1001 etc. The nearest palindromic number will be n-1<\/code> times 9 where n<\/code> is the length of number. For example 1001<\/code>‘s nearest palindrome is 999<\/code>.<\/li>\n
  • Case -4:<\/code> If the number is not a palindrome, and does not come in the above-mentioned cases, then we print out the first half of number as the second half, so abcdef<\/code> is printed as abccba<\/code>.<\/li>\n
  • Case -5:<\/code> If the number is a palindrome, we decrease the central digit by one in one case, and increase by 1 in second. On comparing whichever produces the smallest number we print that case as our result.
    \n
    <\/blockquote>\n<\/li>\n<\/ul>\n
    Example<\/h4>\n
    Let’s discuss above cases with examples:  <\/p>\n
    Case-1: If number is single is digit.<\/em><\/strong>
    \nLets assume number to be 5<\/code>, so we print 1<\/code> number less to it, so 4<\/code> is answer.  <\/p>\n
    Case-2: If number contains only 9 in it.<\/em><\/strong>
    \nLets assume number to be 999<\/code>, so answer would be number+<\/code> 2, Hence 1001<\/code> is our answer.  <\/p>\n
    Case-3: If number contains 0 and 1 only, and number ends with 1.<\/em><\/strong>
    \nLets assume number to be 11<\/code>, so answer would be number-<\/code> 2, Hence 9<\/code> is our answer.  <\/p>\n
    Case-4: If number is not a palindrome.<\/em><\/strong>
    \nLets assume number to be 12344<\/code>, so answer would be FirstHalf<\/code> + middleNumber<\/code> + (FirstHalf<\/code> in<\/code> reversed<\/code> order<\/code>), Hence 12321<\/code> is our answer.  <\/p>\n
    Case-5: If number is palindrome.<\/em><\/strong>
    \nLets assume two numbers here, if number is 12321<\/code>, middle digit is 3<\/code>, so we decrement middle digit by 1<\/code>. So, our answer is 12221<\/code>.  <\/p>\n
    If number is 12121<\/code>, middle digit is 1<\/code>, so we increment middle digit by 1<\/code>. So, our answer is 12221<\/code>.<\/p>\n
    Solutions<\/h3>\n
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