{"id":994,"date":"2020-06-11T10:15:19","date_gmt":"2020-06-11T10:15:19","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=994"},"modified":"2022-03-28T01:16:39","modified_gmt":"2022-03-28T01:16:39","slug":"convert-integer-number-to-roman-number","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/convert-integer-number-to-roman-number\/","title":{"rendered":"Convert Integer number to Roman Number"},"content":{"rendered":"

\"\"<\/p>\n

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

Difficulty Level:<\/h3>\n
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Medium<\/p>\n<\/blockquote>\n

Problem Statement (Simplified):<\/h3>\n
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For a given number between 0 and 4000, print out it’s Roman Form.<\/p>\n<\/blockquote>\n

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

Test Case:<\/h4>\n
Input:\n1\n49\n\nOutput:\nXLIX\n\nExplanation:\nStarting from left we have 4<\/code> at 10th digit place, hence it is 40, 40 can be written as XL in Roman Numerals.\n\nNext, we have 9 at unit digit place which can be written as IX in Roman Numerals.\n\nSo, our final answer is XLIX. <\/code><\/pre>\n
Solving Approach :<\/h3>\n
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1) Roman numerals with their Integer counterpart are as follows :<\/p>\n<\/blockquote>\n
1    -  I\n5    -  V\n10   -  X\n50   -  L\n100  -  C\n500  -  D\n1000 -  M<\/code><\/pre>\n
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2) We scan number from left to right digit by digit and print digit’s respective value depending on its unit place whether 1000th unit place, 100th unit place, 10th unit place, or unit place.<\/p>\n
3) We follow the given chart for printing values according to their place in the number.<\/p>\n<\/blockquote>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Digit<\/th>\nUnit Place<\/th>\nTenth Place<\/th>\nHundredth Place<\/th>\nThousandth Place<\/th>\n<\/tr>\n<\/thead>\n
1<\/td>\nI<\/td>\nX<\/td>\nC<\/td>\nM<\/td>\n<\/tr>\n
2<\/td>\nII<\/td>\nXX<\/td>\nCC<\/td>\nMM<\/td>\n<\/tr>\n
3<\/td>\nIII<\/td>\nXXX<\/td>\nCCC<\/td>\nMMM<\/td>\n<\/tr>\n
4<\/td>\nIV<\/td>\nXL<\/td>\nCD<\/td>\n–<\/td>\n<\/tr>\n
5<\/td>\nV<\/td>\nL<\/td>\nD<\/td>\n–<\/td>\n<\/tr>\n
6<\/td>\nVI<\/td>\nLX<\/td>\nDC<\/td>\n–<\/td>\n<\/tr>\n
7<\/td>\nVII<\/td>\nLXX<\/td>\nDCC<\/td>\n–<\/td>\n<\/tr>\n
8<\/td>\nVIII<\/td>\nLXXX<\/td>\nDCCC<\/td>\n–<\/td>\n<\/tr>\n
9<\/td>\nIX<\/td>\nXC<\/td>\nCM<\/td>\n–<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Example:<\/h4>\n
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    \n
  • Let’s take 2975 as an example, we scan elements from left to right digit by digit.<\/li>\n<\/ul>\n
    \"\"<\/p>\n
      \n
    • \n
      At 1<\/code>st<\/sup> place we have, 2<\/code> at Thousandth digit place, which means we have 2000 in number, so 2000 can be written as MM<\/code> in Roman Numerals.<\/p>\n<\/li>\n
    • \n
      At 2<\/code>nd<\/sup> place we have, 9<\/code> at Hundredth digit place, which means we have 900 in number, so 900 can be written as CM<\/code> in Roman Numerals.<\/p>\n<\/li>\n
    • \n
      At 3<\/code>rd<\/sup> place we have, 7<\/code> at Tenth digit place, which means we have 70 in number, so 70 can be written as LXX<\/code> in Roman Numerals.<\/p>\n<\/li>\n
    • \n
      At last place we have, 5<\/code> at Thousandth digit place, which means we have 5 in number, so 5 can be written as V<\/code> in Roman Numerals.<\/p>\n<\/li>\n
    • \n
      Finally, we concatenate all of them, so 2975 can be written as MMCMLXXV<\/code><\/p>\n<\/li>\n<\/ul>\n<\/blockquote>\n
      Solutions:<\/h3>\n\t\t\t\t\t\t