Dynamic 🧨 Programming 📝 | Part II of X | Subproblem Space & Relationship Expansion 🚀 | LCS & Edit Distance | DSA 🤹🏻 Magic 🧞‍♂️
Longest Common Subsequence | LCS of THEIR - HABIT is (to say) HI

Dynamic 🧨 Programming 📝 | Part II of X | Subproblem Space & Relationship Expansion 🚀 | LCS & Edit Distance | DSA 🤹🏻 Magic 🧞♂️

Krishna R. Krishna R.

Krishna R.

B-Tech IIT Madras | MS Penn State | Ex Intel | Ex CSA Addverb

Published Sep 2, 2024
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𝖳𝗁𝗂𝗌 𝗂𝗌 𝗍𝗁𝖾 𝗌𝖾𝖼𝗈𝗇𝖽 𝗂𝗇 𝖺 𝗦𝗲𝗿𝗶𝗲𝘀 𝗈𝖿 𝖺𝗋𝗍𝗂𝖼𝗅𝖾𝗌 𝗈𝗇 𝗗𝘆𝗻𝗮𝗺𝗶𝗰 𝗣𝗿𝗼𝗴𝗿𝗮𝗺𝗺𝗶𝗻𝗴 𝗍𝗁𝖺𝗍 𝗌𝗍𝖺𝗋𝗍𝖾𝖽 𝗐𝗂𝗍𝗁 𝗍𝗁𝗂𝗌. 𝖠𝗇𝖽 𝗀𝗂𝗏𝖾𝗇 𝗍𝗁𝖺𝗍 𝗐𝖾 𝖼𝗈𝗏𝖾𝗋𝖾𝖽 𝗌𝗈𝗆𝖾 𝗈𝖿 𝗍𝗁𝖾 𝖾𝗅𝖾𝗆𝖾𝗇𝗍𝖺𝗋𝗒 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗂𝗇 𝗍𝗁𝖾 𝖿𝗂𝗋𝗌𝗍 𝗈𝗇𝖾 𝗐𝖾'𝗅𝗅 𝖼𝗋𝖺𝗇𝗄 𝗂𝗍 𝗎𝗉 𝖺 𝗇𝗈𝗍𝖼𝗁 𝗇𝗈𝗐 𝖺𝗇𝖽 𝗌𝗍𝖺𝗋𝗍 𝗅𝗈𝗈𝗄𝗂𝗇𝗀 𝖺𝗍

𝖺. 𝖧𝗈𝗐 𝗍𝗈 𝖾𝗑𝗉𝖺𝗇𝖽 𝗍𝗁𝖾 𝖲𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖲𝗉𝖺𝖼𝖾 𝖺𝗇𝖽 𝖱𝖾𝗅𝖺𝗍𝗂𝗈𝗇𝗌𝗁𝗂𝗉𝗌 𝖻𝗒 𝖺𝖽𝖽𝗂𝗇𝗀 𝖼𝗈𝗇𝗌𝗍𝗋𝖺𝗂𝗇𝗍𝗌 𝗍𝗁𝖺𝗍 𝗁𝖾𝗅𝗉 𝗎𝗌 𝗍𝗋𝖺𝖼𝗄 𝖺𝖽𝖽𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝗌𝗍𝖺𝗍𝖾

𝖻. 𝖲𝗎𝖻𝗌𝗍𝗋𝗂𝗇𝗀𝗌 (𝖺𝗌 𝗌𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌) 𝗂𝗇𝗌𝗍𝖾𝖺𝖽 𝗈𝖿 𝗍𝗁𝖾 𝗃𝗎𝗌𝗍 𝗉𝗋𝖾𝖿𝗂𝗑𝖾𝗌 𝖺𝗇𝖽 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌

𝖳𝗁𝖾𝗌𝖾 𝖺𝗋𝖾 𝗌𝗈𝗆𝖾 𝗈𝖿 𝗍𝗁𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗐𝖾'𝗅𝗅 𝖻𝖾 𝗍𝖺𝗄𝗂𝗇𝗀 𝖺 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗎𝗇𝖽𝖾𝗋 𝖾𝖺𝖼𝗁 𝖼𝖺𝗍𝖾𝗀𝗈𝗋𝗒:

𝟣. 𝗖𝗮𝗿𝘁𝗲𝘀𝗶𝗮𝗻 𝗣𝗿𝗼𝗱𝘂𝗰𝘁 𝗈𝖿 𝖳𝗐𝗈 𝗦𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺 𝗦𝗽𝗮𝗰𝗲𝗌:

𝖺. 𝗦𝘂𝗳𝗳𝗶𝘅𝗲𝘀 𝖷 𝗦𝘂𝗳𝗳𝗶𝘅𝗲𝘀 (𝗈𝗋 𝗣𝗿𝗲𝗳𝗶𝘅𝗲𝘀 𝖷 𝗣𝗿𝗲𝗳𝗶𝘅𝗲𝘀):

𝗂. 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 | 𝖬𝗂𝗇𝗂𝗆𝗂𝗓𝖾 𝗍𝗁𝖾 𝖤𝖽𝗂𝗍 𝖣𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝖻𝖾𝗍𝗐𝖾𝖾𝗇 𝗍𝗐𝗈 𝗌𝗍𝗋𝗂𝗇𝗀𝗌

𝗂𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗖𝗼𝗺𝗺𝗼𝗻 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 / 𝖲𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖠𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 (𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗇𝗈𝗍 𝗏𝖾𝗋𝗒 𝖽𝗂𝖿𝖿𝖾𝗋𝖾𝗇𝗍 𝖿𝗋𝗈𝗆 𝖤𝖽𝗂𝗍 𝖣𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝖺𝗇𝖽 𝗁𝗈𝗐 𝗍𝗁𝖾𝗒 𝖺𝗋𝖾 𝗍𝗐𝗈 𝗌𝗂𝖽𝖾𝗌 𝗈𝖿 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝖼𝗈𝗂𝗇)

𝖻. 𝖨𝗇 𝗅𝖺𝗍𝖾𝗋 𝖺𝗋𝗍𝗂𝖼𝗅𝖾𝗌 𝗐𝖾 𝗐𝗂𝗅𝗅 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗆𝗈𝗋𝖾 𝗏𝖺𝗋𝗂𝖾𝗍𝗒 𝗈𝖿 𝗍𝗁𝗂𝗌 𝗄𝗂𝗇𝖽 𝗈𝖿 𝖾𝗑𝗉𝖺𝗇𝗌𝗂𝗈𝗇 𝖿𝗈𝗋 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝗜𝗻𝘁𝗲𝗴𝗲𝗿 𝖷 𝗦𝘂𝗳𝗳𝗶𝘅𝗲𝘀 𝗂𝗇 𝗍𝗁𝖾 𝖼𝖺𝗌𝖾 𝗈𝖿 𝗥𝗼𝗱 𝗖𝘂𝘁𝘁𝗶𝗻𝗴 𝖺𝗇𝖽 𝗦𝘂𝗯𝘀𝗲𝘁 𝗦𝘂𝗺 𝖺𝗇𝖽 𝗌𝗈 𝗈𝗇. 𝖤𝗏𝖾𝗇𝗍𝗎𝖺𝗅𝗅𝗒 𝗐𝖾 𝗐𝗂𝗅𝗅 𝗌𝗎𝗆𝗆𝖺𝗋𝗂𝗓𝖾 𝖺𝗅𝗅 𝗍𝗁𝖾𝗌𝖾 𝗉𝖺𝗍𝗍𝖾𝗋𝗇𝗌 𝗂𝗇 𝖺 𝖼𝗈𝗇𝖼𝗅𝗎𝖽𝗂𝗇𝗀 𝖺𝗋𝗍𝗂𝖼𝗅𝖾 𝗌𝗈 𝗒𝗈𝗎 𝖼𝖺𝗇 𝗀𝗈 𝖻𝖺𝖼𝗄 𝖺𝗇𝖽 𝗍𝖺𝗄𝖾 𝖺 𝗅𝗈𝗈𝗄 𝖺𝗍 𝖺𝗅𝗅 𝗏𝖺𝗋𝗂𝖺𝗍𝗂𝗈𝗇𝗌 𝖺𝗇𝖽 𝗄𝖾𝖾𝗉 𝗍𝗁𝖾𝗆 𝗂𝗇 𝗆𝗂𝗇𝖽 𝖺𝗌 𝗉𝖺𝗋𝗍 𝗈𝖿 𝗒𝗈𝗎𝗋 𝗍𝗈𝗈𝗅 𝗄𝗂𝗍 𝖿𝗈𝗋 𝖣𝖯.

𝟤. 𝗦𝘂𝗯𝘀𝘁𝗿𝗶𝗻𝗴𝗌

𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗣𝗮𝗹𝗶𝗻𝗱𝗿𝗼𝗺𝗶𝗰 𝗦𝘂𝗯𝘀𝘁𝗿𝗶𝗻𝗴

𝗂𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗣𝗮𝗹𝗶𝗻𝗱𝗿𝗼𝗺𝗶𝗰 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲

𝗂𝗂𝗂. 𝗠𝗮𝘅 𝗦𝘂𝗺 𝗦𝘂𝗯𝗮𝗿𝗿𝗮𝘆

𝗂𝗏. 𝗠𝗮𝘅 𝗣𝗿𝗼𝗱𝘂𝗰𝘁 𝗦𝘂𝗯𝗮𝗿𝗿𝗮𝘆

𝟥. 𝗔𝗱𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗖𝗼𝗻𝘀𝘁𝗿𝗮𝗶𝗻𝘁𝘀 (𝖲𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗦𝗽𝗮𝗰𝗲 𝖺𝗇𝖽 𝗥𝗲𝗹𝗮𝘁𝗶𝗼𝗻𝘀𝗵𝗶𝗽 𝗘𝘅𝗽𝗮𝗻𝘀𝗶𝗼𝗻)

𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 (𝖦𝗈𝗈𝖽) 𝗣𝗮𝗹𝗶𝗻𝗱𝗿𝗼𝗺𝗶𝗰 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝖨𝖨

𝗂𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗜𝗻𝗰𝗿𝗲𝗮𝘀𝗶𝗻𝗴 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲

𝖫𝖾𝗍'𝗌 𝗀𝖾𝗍 𝗍𝗈 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝗌𝗈𝗆𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗇𝗈𝗐.

𝖨. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗖𝗼𝗺𝗺𝗼𝗻 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 (𝖫𝖢𝖲)

𝙇𝘾 𝘔𝘦𝘥𝘪𝘶𝘮 1143. 𝙇𝙤𝙣𝙜𝙚𝙨𝙩 𝘾𝙤𝙢𝙢𝙤𝙣 𝙎𝙪𝙗𝙨𝙚𝙦𝙪𝙚𝙣𝙘𝙚

𝙈𝙚𝙙𝙞𝙪𝙢

𝘎𝘪𝘷𝘦𝘯 𝘵𝘸𝘰 𝘴𝘵𝘳𝘪𝘯𝘨𝘴 𝘵𝘦𝘹𝘵1 𝘢𝘯𝘥 𝘵𝘦𝘹𝘵2, 𝘳𝘦𝘵𝘶𝘳𝘯 𝘵𝘩𝘦 𝘭𝘦𝘯𝘨𝘵𝘩 𝘰𝘧 𝘵𝘩𝘦𝘪𝘳 𝘭𝘰𝘯𝘨𝘦𝘴𝘵 𝘤𝘰𝘮𝘮𝘰𝘯 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦. 𝘐𝘧 𝘵𝘩𝘦𝘳𝘦 𝘪𝘴 𝘯𝘰 𝘤𝘰𝘮𝘮𝘰𝘯 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦, 𝘳𝘦𝘵𝘶𝘳𝘯 0.

𝘈 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦 𝘰𝘧 𝘢 𝘴𝘵𝘳𝘪𝘯𝘨 𝘪𝘴 𝘢 𝘯𝘦𝘸 𝘴𝘵𝘳𝘪𝘯𝘨 𝘨𝘦𝘯𝘦𝘳𝘢𝘵𝘦𝘥 𝘧𝘳𝘰𝘮 𝘵𝘩𝘦 𝘰𝘳𝘪𝘨𝘪𝘯𝘢𝘭 𝘴𝘵𝘳𝘪𝘯𝘨 𝘸𝘪𝘵𝘩 𝘴𝘰𝘮𝘦 𝘤𝘩𝘢𝘳𝘢𝘤𝘵𝘦𝘳𝘴 (𝘤𝘢𝘯 𝘣𝘦 𝘯𝘰𝘯𝘦) 𝘥𝘦𝘭𝘦𝘵𝘦𝘥 𝘸𝘪𝘵𝘩𝘰𝘶𝘵 𝘤𝘩𝘢𝘯𝘨𝘪𝘯𝘨 𝘵𝘩𝘦 𝘳𝘦𝘭𝘢𝘵𝘪𝘷𝘦 𝘰𝘳𝘥𝘦𝘳 𝘰𝘧 𝘵𝘩𝘦 𝘳𝘦𝘮𝘢𝘪𝘯𝘪𝘯𝘨 𝘤𝘩𝘢𝘳𝘢𝘤𝘵𝘦𝘳𝘴.

𝘍𝘰𝘳 𝘦𝘹𝘢𝘮𝘱𝘭𝘦, "𝘢𝘤𝘦" 𝘪𝘴 𝘢 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦 𝘰𝘧 "𝘢𝘣𝘤𝘥𝘦".

𝘈 𝘤𝘰𝘮𝘮𝘰𝘯 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦 𝘰𝘧 𝘵𝘸𝘰 𝘴𝘵𝘳𝘪𝘯𝘨𝘴 𝘪𝘴 𝘢 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦 𝘵𝘩𝘢𝘵 𝘪𝘴 𝘤𝘰𝘮𝘮𝘰𝘯 𝘵𝘰 𝘣𝘰𝘵𝘩 𝘴𝘵𝘳𝘪𝘯𝘨𝘴.

𝙀𝙭𝙖𝙢𝙥𝙡𝙚 1:

𝙄𝙣𝙥𝙪𝙩: 𝘵𝘦𝘹𝘵1 = "𝘢𝘣𝘤𝘥𝘦", 𝘵𝘦𝘹𝘵2 = "𝘢𝘤𝘦"

𝙊𝙪𝙩𝙥𝙪𝙩: 3

𝙀𝙭𝙥𝙡𝙖𝙣𝙖𝙩𝙞𝙤𝙣: 𝘛𝘩𝘦 𝘭𝘰𝘯𝘨𝘦𝘴𝘵 𝘤𝘰𝘮𝘮𝘰𝘯 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦 𝘪𝘴 "𝘢𝘤𝘦" 𝘢𝘯𝘥 𝘪𝘵𝘴 𝘭𝘦𝘯𝘨𝘵𝘩 𝘪𝘴 3.

𝙀𝙭𝙖𝙢𝙥𝙡𝙚 2:

𝙄𝙣𝙥𝙪𝙩: 𝘵𝘦𝘹𝘵1 = "𝘢𝘣𝘤", 𝘵𝘦𝘹𝘵2 = "𝘢𝘣𝘤"

𝙊𝙪𝙩𝙥𝙪𝙩: 3

𝙀𝙭𝙥𝙡𝙖𝙣𝙖𝙩𝙞𝙤𝙣: 𝘛𝘩𝘦 𝘭𝘰𝘯𝘨𝘦𝘴𝘵 𝘤𝘰𝘮𝘮𝘰𝘯 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦 𝘪𝘴 "𝘢𝘣𝘤" 𝘢𝘯𝘥 𝘪𝘵𝘴 𝘭𝘦𝘯𝘨𝘵𝘩 𝘪𝘴 3.

𝙀𝙭𝙖𝙢𝙥𝙡𝙚 3:

𝙄𝙣𝙥𝙪𝙩: 𝘵𝘦𝘹𝘵1 = "𝘢𝘣𝘤", 𝘵𝘦𝘹𝘵2 = "𝘥𝘦𝘧"

𝙊𝙪𝙩𝙥𝙪𝙩: 0

𝙀𝙭𝙥𝙡𝙖𝙣𝙖𝙩𝙞𝙤𝙣: 𝘛𝘩𝘦𝘳𝘦 𝘪𝘴 𝘯𝘰 𝘴𝘶𝘤𝘩 𝘤𝘰𝘮𝘮𝘰𝘯 𝘴𝘶𝘣𝘴𝘦𝘲𝘶𝘦𝘯𝘤𝘦, 𝘴𝘰 𝘵𝘩𝘦 𝘳𝘦𝘴𝘶𝘭𝘵 𝘪𝘴 0.

𝖡𝖾𝖿𝗈𝗋𝖾 𝗐𝖾 𝗉𝗋𝗈𝖼𝖾𝖾𝖽 𝗍𝗈 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝗍𝗁𝗂𝗌 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖺 𝖿𝖾𝗐 𝗋𝖾𝗆𝖺𝗋𝗄𝗌 𝖺𝗋𝖾 𝗂𝗇 𝗈𝗋𝖽𝖾𝗋. 𝖶𝖾 𝗐𝗂𝗅𝗅 𝖻𝖾 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝖺𝗇𝖽 𝗂𝗅𝗅𝗎𝗌𝗍𝗋𝖺𝗍𝗂𝗇𝗀 𝖺 𝗌𝗂𝗀𝗇𝗂𝖿𝗂𝖼𝖺𝗇𝗍 𝗉𝖺𝗋𝗍 𝗈𝖿 𝗍𝗁𝖾 𝗦𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗍𝗁𝖺𝗍 𝖺𝗉𝗉𝗅𝗂𝖾𝗌 𝗍𝗈 𝖻𝗈𝗍𝗁 𝗍𝗁𝖾 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗖𝗼𝗺𝗺𝗼𝗻 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲, 𝖺𝗄𝖺 𝖬𝖺𝗑𝗂𝗆𝗂𝗓𝗂𝗇𝗀 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁, 𝖺𝗇𝖽 𝗍𝗁𝖾 𝗇𝖾𝗑𝗍 𝗈𝗇𝖾 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 (𝖬𝗂𝗇𝗂𝗆𝗂𝗓𝗂𝗇𝗀) 𝗍𝗁𝖾 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲. 𝖱𝖾𝖺𝗌𝗈𝗇 𝖻𝖾𝗂𝗇𝗀 𝗍𝗁𝖺𝗍 𝗍𝗁𝖾𝗌𝖾 𝗍𝗐𝗈 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝖺𝗋𝖾 𝗂𝖽𝖾𝗇𝗍𝗂𝖼𝖺𝗅 𝗆𝖺𝗍𝗁𝖾𝗆𝖺𝗍𝗂𝖼𝖺𝗅𝗅𝗒 𝖾𝗏𝖾𝗇 𝗍𝗁𝗈𝗎𝗀𝗁 𝗍𝗁𝖾 𝖼𝗈𝖽𝖾 𝗆𝗂𝗀𝗁𝗍 𝗅𝗈𝗈𝗄 𝗌𝗈𝗆𝖾𝗐𝗁𝖺𝗍 𝖽𝗂𝖿𝖿𝖾𝗋𝖾𝗇𝗍 𝗈𝗇 𝗍𝗁𝖾 𝗌𝗎𝗋𝖿𝖺𝖼𝖾 𝖻𝗈𝗍𝗁 𝗍𝗁𝖾 𝗂𝖽𝖾𝖺𝗌 𝖺𝗋𝖾 𝖾𝗌𝗌𝖾𝗇𝗍𝗂𝖺𝗅𝗅𝗒 𝗍𝗐𝗈 𝗌𝗂𝖽𝖾𝗌 𝗈𝖿 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝖼𝗈𝗂𝗇 𝗈𝖿 𝗦𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁.

𝖳𝗈 𝗎𝗇𝖽𝖾𝗋𝗌𝗍𝖺𝗇𝖽 𝗍𝗁𝗂𝗌 𝖻𝖾𝗍𝗍𝖾𝗋 𝗅𝖾𝗍'𝗌 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋 𝗍𝗁𝖾 𝖿𝗈𝗅𝗅𝗈𝗐𝗂𝗇𝗀 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝖿𝗈𝗋 𝗈𝗎𝗋 𝗂𝗅𝗅𝗎𝗌𝗍𝗋𝖺𝗍𝗂𝗈𝗇 (𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗂𝗇 𝗍𝗁𝗂𝗌 𝖼𝖺𝗌𝖾 𝖻𝖾𝗂𝗇𝗀 𝖺 𝗌𝗍𝗋𝗂𝗇𝗀 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗄𝗂𝗇𝖽 𝗈𝖿 "𝖺 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗈𝖿 𝖼𝗁𝖺𝗋𝖺𝖼𝗍𝖾𝗋𝗌". 𝖡𝗎𝗍 𝗄𝖾𝖾𝗉 𝗂𝗇 𝗆𝗂𝗇𝖽 𝗍𝗁𝗂𝗌 𝖺𝗉𝗉𝗅𝗂𝖾𝗌 𝗍𝗈 𝖺𝗅𝗆𝗈𝗌𝗍 𝖺𝗇𝗒 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗍𝗁𝖺𝗍 𝗁𝖺𝗌 𝖤𝗊𝗎𝖺𝗍𝖺𝖻𝗅𝖾 𝖾𝗅𝖾𝗆𝖾𝗇𝗍𝗌 𝗐𝗁𝗂𝖼𝗁 𝖾𝗇𝖺𝖻𝗅𝖾𝗌 𝗒𝗈𝗎 𝗍𝗈 𝖽𝗈 𝖺𝗇 𝖾𝗅𝖾𝗆𝖾𝗇𝗍 𝖻𝗒 𝖾𝗅𝖾𝗆𝖾𝗇𝗍 𝖼𝗈𝗆𝗉𝖺𝗋𝗂𝗌𝗈𝗇 𝗈𝖿 𝖾𝗊𝗎𝖺𝗅𝗂𝗍𝗒.).

𝘀𝟣 = 𝗗 𝗜 𝗦 𝗧 𝗔 𝗡 𝗖 𝗘

𝘀𝟤 = 𝗘 𝗗 𝗜 𝗧 𝗜 𝗡 𝗚

𝖶𝗁𝖾𝗇 𝗐𝖾 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗍𝗁𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗍𝗁𝗋𝗈𝗎𝗀𝗁 𝗍𝗁𝖾 𝗅𝖾𝗇𝗌 𝗈𝖿 𝖬𝗂𝗇𝗂𝗆𝗂𝗓𝗂𝗇𝗀 𝖤𝖽𝗂𝗍 𝖣𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗍𝗁𝖾 𝗉𝖾𝗋𝗌𝗉𝖾𝖼𝗍𝗂𝗏𝖾 𝖻𝖾𝖼𝗈𝗆𝖾𝗌 𝗆𝗎𝖼𝗁 𝖼𝗅𝖾𝖺𝗋𝖾𝗋. 𝖬𝗂𝗇𝗂𝗆𝗂𝗓𝗂𝗇𝗀 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗂𝗌 𝖺𝖻𝗈𝗎𝗍 𝗆𝗈𝗋𝗉𝗁𝗂𝗇𝗀 𝗍𝗁𝖾 𝗌𝗍𝗋𝗂𝗇𝗀 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝗂𝗇𝗍𝗈 𝗘𝗗𝗜𝗧𝗜𝗡𝗚 𝗎𝗌𝗂𝗇𝗀 𝗆𝗂𝗇𝗂𝗆𝖺𝗅 𝖾𝖽𝗂𝗍𝗂𝗇𝗀 𝖺𝖼𝗍𝗂𝗈𝗇𝗌 𝗐𝗁𝗂𝖼𝗁 𝖺𝗋𝖾 𝖺𝗅𝗅𝗈𝗐𝖾𝖽 𝗍𝗈 𝖻𝖾 𝗈𝗇𝖾 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗅𝗅𝗈𝗐𝗂𝗇𝗀:

𝖺. 𝗜𝗻𝘀𝗲𝗿𝘁 𝖺 𝖼𝗁𝖺𝗋𝖺𝖼𝗍𝖾𝗋 𝗂𝗇𝗍𝗈 𝗍𝗁𝖾 𝖺𝗋𝗋𝖺𝗒 𝗌𝟣 (𝖺𝗍 𝖺𝗇𝗒 𝗂𝗇𝖽𝖾𝗑)

𝖻. 𝗗𝗲𝗹𝗲𝘁𝗲 𝖺 𝖼𝗁𝖺𝗋𝖺𝖼𝗍𝖾𝗋 𝗂𝗇𝗍𝗈 𝗍𝗁𝖾 𝖺𝗋𝗋𝖺𝗒 𝗌𝟣 (𝖺𝗍 𝖺𝗇𝗒 𝗂𝗇𝖽𝖾𝗑)

𝖼. 𝗥𝗲𝗽𝗹𝗮𝗰𝗲 𝖺 𝖼𝗁𝖺𝗋𝖺𝖼𝗍𝖾𝗋 𝗂𝗇𝗍𝗈 𝗍𝗁𝖾 𝖺𝗋𝗋𝖺𝗒 𝗌𝟣 (𝖺𝗍 𝖺𝗇𝗒 𝗂𝗇𝖽𝖾𝗑)

𝖫𝖾𝗍'𝗌 𝗍𝖺𝗄𝖾 𝖺 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗈𝗇𝖾 𝗉𝗈𝗌𝗌𝗂𝖻𝗅𝖾 𝗈𝗉𝗍𝗂𝗆𝖺𝗅 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 𝗈𝖿 𝗍𝗁𝖾 𝗌𝗍𝗋𝗂𝗇𝗀𝗌 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝖺𝗇𝖽 𝗘𝗗𝗜𝗧𝗜𝗡𝗚 𝖻𝖾𝗅𝗈𝗐 (𝖺𝗅𝗌𝗈 𝗄𝖾𝖾𝗉 𝗂𝗇 𝗆𝗂𝗇𝖽 𝗍𝗁𝖺𝗍 𝗐𝖾 𝗌𝗁𝗈𝗎𝗅𝖽 𝖻𝖾 𝖼𝖺𝗋𝖾𝖿𝗎𝗅 𝗇𝗈𝗍 𝗍𝗈 𝗋𝖾𝖿𝖾𝗋 𝗍𝗈 "𝘁𝗵𝗲" 𝖫𝗈𝗇𝗀𝖾𝗌𝗍 𝖢𝗈𝗆𝗆𝗈𝗇 𝖲𝗎𝖻𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗎𝗇𝗅𝖾𝗌𝗌 𝗐𝖾 𝗁𝖺𝗏𝖾 𝗉𝗋𝗈𝗏𝖾𝗇 𝗍𝗁𝖺𝗍 𝗂𝗍 𝗂𝗌 𝗂𝗇𝖽𝖾𝖾𝖽 𝗎𝗇𝗂𝗊𝗎𝖾. 𝖨𝗇 𝗆𝗈𝗌𝗍 𝖼𝖺𝗌𝖾𝗌 𝗍𝗁𝖾𝗋𝖾 𝖺𝗋𝖾 𝗆𝖺𝗇𝗒 𝗅𝗈𝗇𝗀𝖾𝗌𝗍 𝖼𝗈𝗆𝗆𝗈𝗇 𝗌𝗎𝖻𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝖻𝖾𝗍𝗐𝖾𝖾𝗇 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝖺𝗇𝖽 𝗌𝗈 𝗐𝖾 𝗌𝗁𝗈𝗎𝗅𝖽 𝗋𝖾𝖿𝖾𝗋 𝗍𝗈 𝗂𝗍 𝖺𝗇𝖽 𝗎𝗇𝖽𝖾𝗋𝗌𝗍𝖺𝗇𝖽 𝗂𝗍 𝗂𝗇 𝗈𝗎𝗋 𝗆𝗂𝗇𝖽 𝗍𝗈 𝖻𝖾 "𝗮" 𝖫𝗈𝗇𝗀𝖾𝗌𝗍 𝖢𝗈𝗆𝗆𝗈𝗇 𝖲𝗎𝖻𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾).

Alignment of Sequences | DISTANCE and EDITING

𝖸𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝗈𝗇𝖾 𝗈𝖿 𝗍𝗁𝖾 𝗈𝗉𝗍𝗂𝗆𝖺𝗅 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍𝗌 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝖺𝖻𝗈𝗏𝖾. 𝖳𝗁𝖾 𝗐𝖺𝗒 𝗍𝗈 𝗋𝖾𝖺𝖽 / 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍 𝗂𝗍 𝗂𝗌 𝖺𝗌 𝖿𝗈𝗅𝗅𝗈𝗐𝗌:

𝗂. 𝗜𝗻𝘀𝗲𝗿𝘁: 𝖺𝗇 𝗘 𝖺𝗍 𝗂𝗇𝖽𝖾𝗑 𝟢 𝗈𝖿 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘

𝗂𝗂. 𝗠𝗮𝘁𝗰𝗵: 𝖺𝗇 𝗂𝗇-𝗉𝗅𝖺𝖼𝖾 𝗇𝖺𝗍𝗎𝗋𝖺𝗅 𝗆𝖺𝗍𝖼𝗁 𝗈𝖿 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋𝗌 𝗗 𝖺𝗇𝖽 𝗜 (𝗌𝗈 𝗌𝗄𝗂𝗉 𝖺𝗁𝖾𝖺𝖽 - 𝗇𝗈 𝖺𝖼𝗍𝗂𝗈𝗇 𝗇𝖾𝖾𝖽𝖾𝖽)

𝗂𝗂𝗂. 𝗗𝗲𝗹𝗲𝘁𝗲: 𝖼𝗁𝖺𝗋 𝗦 𝗂𝗇 𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖺𝗍 𝗂𝗇𝖽𝖾𝗑 𝟥

𝗂𝗏. 𝗠𝗮𝘁𝗰𝗵: 𝖺𝗇 𝗂𝗇-𝗉𝗅𝖺𝖼𝖾 𝗇𝖺𝗍𝗎𝗋𝖺𝗅 𝗆𝖺𝗍𝖼𝗁 𝗈𝖿 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋 𝗧 (𝗌𝗈 𝗌𝗄𝗂𝗉 𝖺𝗁𝖾𝖺𝖽 - 𝗇𝗈 𝖺𝖼𝗍𝗂𝗈𝗇 𝗇𝖾𝖾𝖽𝖾𝖽)

𝗏. 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵: 𝗈𝖿 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋𝗌 𝗔 𝖺𝗇𝖽 𝗜 𝖺𝗍 𝗂𝗇𝖽𝖾𝗑 𝟧 (𝗐𝖾 𝖼𝖺𝗇 𝗀𝖾𝗍 𝖺𝗋𝗈𝗎𝗇𝖽 𝗍𝗁𝗂𝗌 𝖻𝗒 𝗥𝗲𝗽𝗹𝗮𝗰𝗲-𝗂𝗇𝗀 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋 𝗔 𝗐𝗂𝗍𝗁 𝗜 𝗂𝗇 𝗍𝗁𝖾 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗐𝗁𝖾𝗋𝖾𝖺𝗌 𝖿𝗈𝗋 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 (𝖫𝖢𝖲) 𝗐𝖾 𝗃𝗎𝗌𝗍 𝖽𝗈𝗇'𝗍 𝖼𝗈𝗎𝗇𝗍 (𝗂𝗀𝗇𝗈𝗋𝖾) 𝗍𝗁𝖾 𝗆𝗂𝗌𝗆𝖺𝗍𝖼𝗁𝖾𝗌)

𝗏𝗂. 𝗠𝗮𝘁𝗰𝗵: 𝖺𝗇 𝗂𝗇-𝗉𝗅𝖺𝖼𝖾 𝗇𝖺𝗍𝗎𝗋𝖺𝗅 𝗆𝖺𝗍𝖼𝗁 𝗈𝖿 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋 𝖭 (𝗌𝗈 𝗌𝗄𝗂𝗉 𝖺𝗁𝖾𝖺𝖽 - 𝗇𝗈 𝖺𝖼𝗍𝗂𝗈𝗇 𝗇𝖾𝖾𝖽𝖾𝖽)

𝗏𝗂𝗂. 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵: 𝗈𝖿 𝖼𝗁𝖺𝗋𝗌 𝗖 𝖺𝗇𝖽 𝗚 𝖺𝗍 𝗂𝗇𝖽𝖾𝗑 𝟩 (𝗐𝖾 𝖼𝖺𝗇 𝗀𝖾𝗍 𝖺𝗋𝗈𝗎𝗇𝖽 𝗍𝗁𝗂𝗌 𝖻𝗒 𝗥𝗲𝗽𝗹𝗮𝗰𝗲-𝗂𝗇𝗀 𝗖 𝗐𝗂𝗍𝗁 𝗚 𝗂𝗇 𝗍𝗁𝖾 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗐𝗁𝖾𝗋𝖾𝖺𝗌 𝖿𝗈𝗋 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 (𝖫𝖢𝖲) 𝗐𝖾 𝗃𝗎𝗌𝗍 𝖽𝗈𝗇'𝗍 𝖼𝗈𝗎𝗇𝗍 𝗆𝗂𝗌𝗆𝖺𝗍𝖼𝗁𝖾𝗌)

𝗏𝗂𝗂𝗂. 𝗗𝗲𝗹𝗲𝘁𝗲: 𝖼𝗁𝖺𝗋 𝗘 𝗂𝗇 𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖺𝗍 𝗂𝗇𝖽𝖾𝗑 𝟪

Editing of Sequences | DISTANCE and EDITING

𝖠𝗇𝖽 𝖺𝗌 𝗒𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝖾𝗏𝖾𝗋𝗒 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵 𝗂𝗇 𝗍𝗁𝖾 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁 𝖼𝖺𝗇 𝖻𝖾 𝖾𝗅𝗂𝗆𝗂𝗇𝖺𝗍𝖾𝖽 𝖻𝗒 𝗍𝗁𝖾 𝗈𝗉𝖾𝗋𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗥𝗲𝗽𝗹𝗮𝗰𝗲-𝗆𝖾𝗇𝗍. 𝖶𝗁𝖾𝗋𝖾𝗏𝖾𝗋 𝗍𝗁𝖾𝗋𝖾 𝗂𝗌 𝖺 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵 𝗒𝗈𝗎 𝗥𝗲𝗽𝗹𝗮𝗰𝗲 𝗍𝗁𝖾 𝗆𝗂𝗌𝗆𝖺𝗍𝖼𝗁𝗂𝗇𝗀 𝖼𝗁𝖺𝗋 𝗐𝗂𝗍𝗁 𝗍𝗁𝖺𝗍 𝗈𝖿 𝗍𝗁𝖾 𝖼𝗈𝗋𝗋𝖾𝗌𝗉𝗈𝗇𝖽𝗂𝗇𝗀 𝖼𝗁𝖺𝗋 𝗂𝗇 𝗍𝗁𝖾 𝗈𝗍𝗁𝖾𝗋 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾. 𝖨𝗇 𝗍𝗁𝗂𝗌 𝖼𝖺𝗌𝖾 𝗐𝖾 𝗥𝗲𝗽𝗹𝗮𝗰𝗲 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋𝗌 𝗔 𝖺𝗇𝖽 𝗖 𝗐𝗂𝗍𝗁 𝗜 𝖺𝗇𝖽 𝗚 𝗋𝖾𝗌𝗉𝖾𝖼𝗍𝗂𝗏𝖾𝗅𝗒. 𝖠𝗇𝖽 𝗐𝗁𝖺𝗍 𝗒𝗈𝗎 𝗍𝗁𝖾𝗇 𝗀𝖾𝗍 𝗂𝗌 𝖺 𝖼𝗈𝗆𝗉𝗅𝖾𝗍𝖾 𝗍𝗋𝖺𝗇𝗌𝖿𝗈𝗋𝗆𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝗂𝗇𝗍𝗈 𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗘𝗗𝗜𝗧𝗜𝗡𝗚.

Minimizing Edit Distance is the same as Maximizing Alignment

𝖠𝗇𝖽 𝗐𝖾'𝗅𝗅 𝗌𝗈𝗅𝗏𝖾 (𝘮𝘪𝘯𝘪𝘮𝘪𝘻𝘪𝘯𝘨 𝗍𝗁𝖾) 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗇𝖾𝗑𝗍 𝖻𝗎𝗍 𝗂𝗍'𝗌 𝗎𝗌𝖾𝖿𝗎𝗅 𝗍𝗈 𝗇𝗈𝗍𝖾 𝗍𝗁𝖺𝗍 𝖻𝗒 𝘔𝘪𝘯𝘪𝘮𝘪𝘻𝘪𝘯𝘨 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 𝖻𝖾𝗍𝗐𝖾𝖾𝗇 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝖺𝗇𝖽 𝗘𝗗𝗜𝗧𝗜𝗡𝗚 𝗐𝖾 𝖺𝗅𝗌𝗈 𝖾𝗇𝖽 𝗎𝗉 𝗈𝖻𝗍𝖺𝗂𝗇𝗂𝗇𝗀 𝗍𝗁𝖾 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗖𝗼𝗺𝗺𝗼𝗻 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝖺𝗄𝖺 𝗗𝗜𝗧𝗡 𝗂𝗇 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝖺𝗇𝖽 𝗏𝗂𝖼𝖾 𝗏𝖾𝗋𝗌𝖺. 𝖠𝗇𝖽 𝗍𝗁𝗂𝗌 𝗂𝗌 𝗇𝗈𝗍 𝖺 𝖼𝗈𝗂𝗇𝖼𝗂𝖽𝖾𝗇𝖼𝖾 𝖺𝗌 𝘔𝘪𝘯𝘪𝘮𝘪𝘻𝘪𝘯𝘨 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 𝗂𝗌 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝖺𝗌 𝘔𝘢𝘹𝘪𝘮𝘪𝘻𝘪𝘯𝘨 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁. 𝖳𝗈 𝗌𝖾𝖾 𝗍𝗁𝖺𝗍 𝗅𝖾𝗍'𝗌 𝖼𝗈𝗎𝗇𝗍 𝗍𝗁𝖾 𝗍𝗈𝗍𝖺𝗅 𝗇𝗎𝗆𝖻𝖾𝗋 𝗈𝖿 𝗌𝗒𝗆𝖻𝗈𝗅𝗌 𝗂𝗇 𝗍𝗁𝖾 𝖿𝗂𝗇𝖺𝗅 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝗍𝗋𝗂𝗇𝗀𝗌:

𝖬𝗈𝗋𝖾 𝗀𝖾𝗇𝖾𝗋𝖺𝗅𝗅𝗒 𝗍𝗁𝖾 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁 𝗦𝗰𝗼𝗿𝗲 𝗂𝗌 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌:

Alignment Score Definition | Maximizing Alignment is the Same as Minimizing Edit Distance

𝖠𝗇𝖽 𝗍𝗁𝖾 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗖𝗼𝗺𝗺𝗼𝗻 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗃𝗎𝗌𝗍 𝖼𝗈𝗎𝗇𝗍𝗌 𝗍𝗁𝖾 𝗇𝗎𝗆𝖻𝖾𝗋 𝗈𝖿 𝗆𝖺𝗍𝖼𝗁𝖾𝗌 𝗂𝗇 𝖺𝗇 𝗈𝗉𝗍𝗂𝗆𝖺𝗅 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍.

𝖠𝗉𝗉𝗅𝗒𝗂𝗇𝗀 𝗍𝗁𝖾 𝗦𝗥𝗧𝗕𝗢𝗧 𝘍𝘳𝘢𝘮𝘦𝘸𝘰𝘳𝘬

𝟣. 𝗦𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀

𝖢𝗅𝖾𝖺𝗋𝗅𝗒 𝗐𝖾 𝖼𝖺𝗇'𝗍 𝗐𝗈𝗋𝗄 𝗐𝗂𝗍𝗁 𝖺 𝗌𝗂𝗇𝗀𝗅𝖾 𝗌𝗎𝖿𝖿𝗂𝗑 𝖺𝗌 𝗐𝖾 𝗁𝖺𝗏𝖾 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗐𝖾 𝗇𝖾𝖾𝖽 𝗂𝗇𝖽𝖾𝗉𝖾𝗇𝖽𝖾𝗇𝗍 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗂𝗇 𝖾𝖺𝖼𝗁 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝗍𝗈 𝖿𝗈𝗋𝗆𝗎𝗅𝖺𝗍𝖾 𝗈𝗎𝗋 𝗌𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌. 𝖳𝗈 𝗌𝗍𝖺𝗍𝖾 𝗈𝗎𝗋 𝗌𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝗐𝖾 𝗇𝖾𝖾𝖽 𝗍𝗈 𝖼𝖺𝗅𝖼𝗎𝗅𝖺𝗍𝖾

𝗟 [ 𝗶 ][ 𝗷 ] = 𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝖺 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗖𝗼𝗺𝗺𝗼𝗻 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝗈𝖿 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗔[𝗶 : ] 𝖺𝗇𝖽 𝗕[𝗷 : ]

𝗟 [ 𝗶 ][ 𝗷 ] = 𝖫𝖢𝖲(𝗔[𝗶 : ], 𝗕[𝗷 : ])

𝖿𝗈𝗋

𝟢 <= 𝗶 <= |𝗔| (𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝗔 = 𝗺) 𝖺𝗇𝖽

𝟢 <= 𝗷 <= |𝗕| (𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝗕 = 𝗻)

Longest Common Subsequence | Subproblems | Suffixes of A X Suffixes of B
Longest Common Subsequence | Relationship | Possibility | A[i] and B[j] Match
Longest Common Subsequence | Relationship | Possibility | A[i] and B[j] Mismatch | B[j] Present
Longest Common Subsequence | Relationship | Possibility | A[i] and B[j] Mismatch | A[i] Present
Longest Common Subsequence | Relationships | Putting it all Together

𝟤. 𝗥𝗲𝗹𝗮𝘁𝗶𝗼𝗻𝘀𝗵𝗶𝗽𝘀

𝖠𝗌 𝗒𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝗂𝗇 𝗍𝗁𝖾 𝖿𝗂𝗀𝗎𝗋𝖾𝗌 𝖺𝖻𝗈𝗏𝖾:

𝖨𝗇 𝗀𝖾𝗇𝖾𝗋𝖺𝗅 𝗍𝗐𝗈 𝗉𝗈𝗌𝗌𝗂𝖻𝗂𝗅𝗂𝗍𝗂𝖾𝗌 𝖺𝗋𝗂𝗌𝖾:

𝖺. 𝖳𝗁𝖾 𝖼𝗁𝖺𝗋𝗌 𝗔[𝗂] 𝖺𝗇𝖽 𝗕[𝗃] 𝗆𝗂𝗌𝗆𝖺𝗍𝖼𝗁 (𝗐𝗁𝖾𝗇 𝗍𝗁𝗂𝗌 𝗁𝖺𝗉𝗉𝖾𝗇𝗌 𝗍𝗁𝖾𝗒 𝖼𝖺𝗇'𝗍 𝖻𝗈𝗍𝗁 𝖻𝖾 𝗉𝗋𝖾𝗌𝖾𝗇𝗍 𝗂𝗇 𝗍𝗁𝖾 𝗟𝗖𝗦 𝖺𝗇𝖽 𝗍𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗍𝗁𝗂𝗌 𝖻𝗂𝖿𝗎𝗋𝖼𝖺𝗍𝖾𝗌 𝗂𝗇𝗍𝗈 𝗍𝗐𝗈 𝗌𝗎𝖻𝖼𝖺𝗌𝖾𝗌):

𝗂. 𝗔[𝗂] 𝗂𝗌 𝗉𝗋𝖾𝗌𝖾𝗇𝗍 𝖺𝗇𝖽 𝗕[𝗃] 𝗂𝗌 𝖺𝖻𝗌𝖾𝗇𝗍 𝖿𝗋𝗈𝗆 𝗍𝗁𝖾 𝗟𝗖𝗦

𝗂𝗂. 𝗕[𝗃] 𝗂𝗌 𝗉𝗋𝖾𝗌𝖾𝗇𝗍 𝖺𝗇𝖽 𝗔[𝗂] 𝗂𝗌 𝖺𝖻𝗌𝖾𝗇𝗍 𝖿𝗋𝗈𝗆 𝗍𝗁𝖾 𝗟𝗖𝗦

𝖶𝖾 𝖽𝗈𝗇'𝗍 𝗄𝗇𝗈𝗐 𝗐𝗁𝗂𝖼𝗁 𝗈𝗇𝖾 𝗂𝗌 𝗋𝗂𝗀𝗁𝗍 𝗌𝗈 𝗐𝖾 𝗃𝗎𝗌𝗍 𝗯𝗿𝘂𝘁𝗲 𝗳𝗼𝗿𝗰𝗲 𝗐𝗂𝗍𝗁 𝖻𝗈𝗍𝗁 𝗍𝗁𝖾 𝗈𝗉𝗍𝗂𝗈𝗇𝗌 𝖺𝗇𝖽 𝗍𝖺𝗄𝖾 𝗍𝗁𝖾 𝗅𝗈𝗇𝗀𝖾𝗌𝗍 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈:

Longest Common Subsequence | Relationship | Possibility | A[i] and B[j] don't Match

𝖻. 𝖳𝗁𝖾 𝖼𝗁𝖺𝗋𝗌 𝗔[𝗂] 𝖺𝗇𝖽 𝗕[𝗃] 𝗆𝖺𝗍𝖼𝗁 (𝖺𝗇𝖽 𝗌𝗈 𝗍𝗁𝖾𝗒 𝗆𝗂𝗀𝗁𝗍 𝖻𝗈𝗍𝗁 𝖺𝗌 𝗐𝖾𝗅𝗅 𝖻𝖾 𝗂𝗇 𝗍𝗁𝖾 𝗟𝗖𝗦):

Longest Common Subsequence | Relationship | Possibility | A[i] and B[j] Match

𝖯𝗎𝗍𝗍𝗂𝗇𝗀 𝗂𝗍 𝖺𝗅𝗅 𝗍𝗈𝗀𝖾𝗍𝗁𝖾𝗋

Longest Common Subsequence | Subproblem Relationships

𝟥. 𝗧𝗼𝗽𝗼𝗹𝗼𝗴𝗶𝗰𝗮𝗹 𝗢𝗿𝗱𝗲𝗿

𝖲𝗂𝗇𝖼𝖾 𝗐𝖾 𝖽𝗈𝗇'𝗍 𝗐𝖺𝗇𝗍 𝗍𝗈 𝗂𝗇𝖼𝗎𝗋 𝗍𝗁𝖾 𝖼𝗈𝗌𝗍 𝗈𝖿 𝗍𝗈𝗉 𝖽𝗈𝗐𝗇 𝗋𝖾𝖼𝗎𝗋𝗌𝗂𝗈𝗇 𝗐𝖾 𝗌𝗈𝗅𝗏𝖾 𝖿𝗈𝗋 𝗌𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝖻𝗈𝗍𝗍𝗈𝗆-𝗎𝗉 𝗂𝗇 𝗍𝗁𝖾 𝗂𝗇𝖼𝗋𝖾𝖺𝗌𝗂𝗇𝗀 𝗌𝗂𝗓𝖾 𝗈𝖿 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗂.𝖾.,

𝗶 𝗀𝗈𝗂𝗇𝗀 𝖿𝗋𝗈𝗆 [𝗺 -> 𝟢] 𝖺𝗇𝖽

𝗷 𝗀𝗈𝗂𝗇𝗀 𝖿𝗋𝗈𝗆 [𝗻 -> 𝟢]

𝟦. 𝗕𝗮𝘀𝗲 𝗖𝗮𝘀𝗲(𝗌)

𝖳𝗁𝖾 𝖡𝖺𝗌𝖾 𝖢𝖺𝗌𝖾 𝗂𝗌 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝗐𝗁𝖾𝗇 𝖾𝗂𝗍𝗁𝖾𝗋 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝖺𝗋𝖾 𝖾𝗆𝗉𝗍𝗒

𝗟[ 𝗺 ][ 𝗷 ] = 𝗟[ 𝗶 ][ 𝗻 ] = 𝟢 𝖿𝗈𝗋 𝖺𝗅𝗅 𝟢 <= 𝗶 <= 𝗺 , 𝟢 <= 𝗷 <= 𝗻

𝟧. 𝗢𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗹𝗲𝗺

𝖳𝗁𝖾 𝗈𝗋𝗂𝗀𝗂𝗇𝖺𝗅 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝗋𝖾𝗊𝗎𝗂𝗋𝖾𝗌 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝖿𝗈𝗋 𝗍𝗁𝖾 𝗟𝗖𝗦 𝗈𝖿 𝖻𝗈𝗍𝗁 𝗍𝗁𝖾 𝖾𝗇𝗍𝗂𝗋𝖾 𝗌𝗍𝗋𝗂𝗇𝗀𝗌 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗟[ 𝟢 ][ 𝟢 ]

𝟨. 𝗧𝗶𝗺𝗲 𝖢𝗈𝗆𝗉𝗅𝖾𝗑𝗂𝗍𝗒

𝖲𝗂𝗇𝖼𝖾 𝗐𝖾 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋 𝖺𝗅𝗅 𝗉𝖺𝗂𝗋𝗌 𝗈𝖿 𝗂𝗇𝖽𝖾𝗉𝖾𝗇𝖽𝖾𝗇𝗍 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗍𝗁𝖾

𝗻𝘂𝗺𝗯𝗲𝗿 𝗈𝖿 𝘀𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀 ~ 𝗢(𝗻^𝟤)

𝘁𝗶𝗺𝗲 𝗉𝖾𝗋 𝘀𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺 𝗂𝗌 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝖼𝗈𝗇𝗌𝗍𝖺𝗇𝗍 𝗍𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗢(𝟣)

𝗍𝗁𝖾 𝗈𝗏𝖾𝗋𝖺𝗅𝗅 𝗍𝗂𝗆𝖾 𝖼𝗈𝗆𝗉𝗅𝖾𝗑𝗂𝗍𝗒 𝗍𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗂𝗌 𝗢(𝗻^𝟤)

𝖫𝖾𝗍'𝗌 𝗌𝖾𝖾 𝗐𝗁𝖺𝗍 𝗍𝗁𝗂𝗌 𝗅𝗈𝗈𝗄𝗌 𝗅𝗂𝗄𝖾 𝗂𝗇 𝖼𝗈𝖽𝖾:

Longest Common Subsequence | Solution

𝖭𝗈𝗍𝖾 𝗍𝗁𝖺𝗍 𝗍𝗁𝖾 𝗇𝖺𝗍𝗎𝗋𝖺𝗅 𝖼𝗁𝗈𝗂𝖼𝖾 𝖿𝗈𝗋 𝗍𝗁𝖾 𝘀𝗲𝗻𝘁𝗶𝗻𝗲𝗹 / 𝗶𝗻𝗶𝘁𝗶𝗮𝗹 𝗏𝖺𝗅𝗎𝖾 𝗂𝗇 𝗆𝗈𝗌𝗍 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗍𝗈 𝖼𝗈𝗆𝗉𝗎𝗍𝖾 𝗺𝗮𝘅 𝗎𝗌𝗎𝖺𝗅𝗅𝗒 𝗂𝗌 -𝗶𝗻𝗳 (𝗈𝗋 𝗍𝗁𝖾 𝗆𝗂𝗇 𝗏𝖺𝗅𝗎𝖾 𝗋𝖾𝗉𝗋𝖾𝗌𝖾𝗇𝗍𝖺𝖻𝗅𝖾 𝖻𝗒 𝗍𝗁𝖾 𝖽𝖺𝗍𝖺𝗍𝗒𝗉𝖾 𝗎𝗌𝖾𝖽 𝗍𝗈 𝖽𝖾𝗇𝗈𝗍𝖾 𝗍𝗁𝖾 𝗏𝖺𝗅𝗎𝖾) 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗋𝖾𝗉𝗋𝖾𝗌𝖾𝗇𝗍𝖾𝖽 𝖻𝗒 𝗳𝗹𝗼𝗮𝘁('-𝗶𝗻𝗳') 𝗂𝗇 𝖯𝗒𝗍𝗁𝗈𝗇. 𝖡𝗎𝗍 𝗐𝖾 𝖼𝖺𝗇 𝗀𝖾𝗍 𝖺𝗐𝖺𝗒 𝗐𝗂𝗍𝗁 𝗃𝗎𝗌𝗍 𝗂𝗇𝗂𝗍𝗂𝖺𝗅𝗂𝗓𝗂𝗇𝗀 𝗐𝗂𝗍𝗁 𝟬 𝖾𝗏𝖾𝗋𝗒𝗐𝗁𝖾𝗋𝖾 𝗂𝗇 𝗍𝗁𝗂𝗌 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖻𝖾𝖼𝖺𝗎𝗌𝖾 𝗍𝗁𝖾 𝗌𝗆𝖺𝗅𝗅𝖾𝗌𝗍 𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝖺𝗇𝗒 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖼𝖺𝗇 𝗈𝗇𝗅𝗒 𝖻𝖾 𝟢.

𝟩. 𝗕 (𝖼𝖺𝗇 𝖺𝗅𝗌𝗈 𝖻𝖾) 𝖿𝗈𝗋 𝗕𝗮𝗰𝗸𝘁𝗿𝗮𝗰𝗸𝗶𝗻𝗴 (𝖳𝗋𝖺𝖼𝗂𝗇𝗀) 𝖯𝖺𝗋𝖾𝗇𝗍 𝗣𝗼𝗶𝗻𝘁𝗲𝗿𝘀

𝖠𝗌 𝗐𝖾 𝗁𝖺𝗏𝖾 𝖺𝗅𝗋𝖾𝖺𝖽𝗒 𝗆𝖾𝗇𝗍𝗂𝗈𝗇𝖾𝖽 𝗍𝗁𝖺𝗍 𝗆𝗈𝗌𝗍 𝗗𝘆𝗻𝗮𝗺𝗶𝗰 𝗣𝗿𝗼𝗴𝗿𝗮𝗺𝗺𝗶𝗻𝗴 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝖼𝖺𝗇 𝖻𝖾 𝗿𝗲𝗱𝘂𝗰𝗲𝗱 𝗍𝗈 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝗦𝗵𝗼𝗿𝘁𝗲𝘀𝘁 𝗣𝗮𝘁𝗵𝗌 𝗂𝗇 𝖺 𝗗𝗶𝗿𝗲𝗰𝘁𝗲𝗱 𝗔𝗰𝘆𝗰𝗹𝗶𝗰 𝗚𝗿𝗮𝗽𝗵 (𝖣𝖠𝖦) 𝗈𝖿 𝗦𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀. 𝖨𝗇 𝖺 𝗅𝖺𝗋𝗀𝖾 𝗌𝗎𝖻𝗌𝖾𝗍 𝗈𝖿 𝗍𝗁𝖾𝗌𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗐𝖾 𝗆𝗂𝗀𝗁𝗍 𝗇𝗈𝗍 𝗃𝗎𝗌𝗍 𝖻𝖾 𝗂𝗇𝗍𝖾𝗋𝖾𝗌𝗍𝖾𝖽 𝗂𝗇 𝖼𝗈𝗆𝗉𝗎𝗍𝗂𝗇𝗀 𝗍𝗁𝖾 𝗉𝗋𝗈𝗉𝖾𝗋𝗍𝗒 𝗈𝖿 𝖺 𝖲𝗁𝗈𝗋𝗍𝖾𝗌𝗍 𝖯𝖺𝗍𝗁 (𝗅𝗂𝗄𝖾 𝗂𝗍𝗌 𝗅𝖾𝗇𝗀𝗍𝗁) 𝗂𝗇 𝖺 𝖣𝖠𝖦 𝖻𝗎𝗍 𝖺𝗅𝗌𝗈 𝗍𝗁𝖾 𝖯𝖺𝗍𝗁 𝗂𝗍𝗌𝖾𝗅𝖿. 𝖠𝗇𝖽 𝖺 𝖼𝗈𝗆𝗉𝗅𝖾𝗍𝖾 𝗆𝖾𝗆𝗈 𝗍𝖺𝖻𝗅𝖾 𝗂𝖿 𝗆𝖺𝗂𝗇𝗍𝖺𝗂𝗇𝖾𝖽 𝗁𝖺𝗌 𝖾𝗇𝗈𝗎𝗀𝗁 𝗂𝗇𝖿𝗈𝗋𝗆𝖺𝗍𝗂𝗈𝗇 𝗍𝗈 𝗌𝗇𝗂𝖿𝖿 𝗍𝗁𝗂𝗌 𝗈𝗎𝗍 𝗈𝖿 𝗂𝗍𝗌𝖾𝗅𝖿. 𝖫𝖾𝗍'𝗌 𝗌𝖾𝖾 𝗁𝗈𝗐 𝗐𝖾 𝖼𝖺𝗇 𝗀𝗈 𝖺𝖻𝗈𝗎𝗍 𝖽𝗈𝗂𝗇𝗀 𝗍𝗁𝖺𝗍 𝗂𝗇 𝗍𝗁𝗂𝗌 𝖲𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖠𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖺𝗄𝖺 𝗟𝗖𝗦.

𝖶𝖾 𝗍𝖺𝗄𝖾 𝗍𝗁𝖾 𝗌𝗆𝖺𝗅𝗅𝖾𝗋 𝖺𝗇𝖽 𝗌𝗂𝗆𝗉𝗅𝖾𝗋 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝗈𝖿 𝗍𝗐𝗈 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌 𝗧𝗛𝗘𝗜𝗥 𝖺𝗇𝖽 𝗛𝗔𝗕𝗜𝗧. 𝖳𝗁𝖾 𝖫𝗈𝗇𝗀𝖾𝗌𝗍 𝖢𝗈𝗆𝗆𝗈𝗇 𝖲𝗎𝖻𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗈𝖿 𝗧𝗛𝗘𝗜𝗥 - 𝗛𝗔𝗕𝗜𝗧 𝗂𝗌 (𝘵𝘰 𝘴𝘢𝘺) 𝗛𝗜. 𝖸𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝗍𝗁𝖾 𝖾𝗇𝗍𝗂𝗋𝖾 𝗗𝗔𝗚 𝖺𝗇𝖽 𝗍𝗁𝖾 𝗆𝖾𝗆𝗈 𝗍𝖺𝖻𝗅𝖾 𝗋𝖾𝗉𝗋𝖾𝗌𝖾𝗇𝗍𝖾𝖽 𝗂𝗇 𝗍𝗁𝖾 𝖿𝗂𝗀𝗎𝗋𝖾 𝖻𝖾𝗅𝗈𝗐. 𝖳𝗁𝖾 𝗏𝖺𝗅𝗎𝖾𝗌 𝗂𝗇𝗌𝗂𝖽𝖾 𝗍𝗁𝖾 𝗇𝗈𝖽𝖾𝗌 𝖺𝗋𝖾 𝗍𝗁𝖾 𝖫𝖢𝖲 𝗏𝖺𝗅𝗎𝖾 𝖿𝗈𝗋 𝗍𝗁𝖺𝗍 𝗌𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆. 𝖥𝗈𝗋 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝗟[ 𝟢 ][ 𝟣 ] = 𝟤. 𝖶𝗁𝗂𝖼𝗁 𝗂𝗌 𝗍𝗈 𝗌𝖺𝗒 𝗍𝗁𝖾 𝗟𝗖𝗦 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗛𝗘𝗜𝗥 𝖺𝗇𝖽 𝗛𝗔𝗕𝗜𝗧 𝗂𝗌 𝟮 (𝗌𝗉𝖾𝖼𝗂𝖿𝗂𝖼𝖺𝗅𝗅𝗒 𝗍𝗁𝖾 𝖼𝗈𝗆𝗆𝗈𝗇 𝗌𝗎𝖻𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗂𝗌 𝗛𝗜)

𝖳𝗁𝖾 𝗄𝖾𝗒 𝗍𝗈 𝗎𝗇𝖽𝖾𝗋𝗌𝗍𝖺𝗇𝖽𝗂𝗇𝗀 𝗍𝗋𝖺𝖼𝗂𝗇𝗀 𝗈𝖿 𝗉𝖺𝗋𝖾𝗇𝗍 𝗉𝗈𝗂𝗇𝗍𝖾𝗋𝗌 𝗂𝗌 𝗂𝗇 𝗋𝖾𝖺𝗅𝗂𝗓𝗂𝗇𝗀 𝗍𝗁𝖺𝗍 𝖺𝗍 𝖾𝖺𝖼𝗁 𝗌𝗍𝖾𝗉 𝗐𝖾 𝖼𝗁𝗈𝗈𝗌𝖾 𝖺 𝗉𝗈𝗌𝗌𝗂𝖻𝗂𝗅𝗂𝗍𝗒, 𝖻𝗋𝖾𝖺𝗄𝗂𝗇𝗀 𝗍𝗂𝖾𝗌, 𝗂𝖿 𝗍𝗁𝖾𝗒 𝖾𝗑𝗂𝗌𝗍, 𝖺𝗋𝖻𝗂𝗍𝗋𝖺𝗋𝗂𝗅𝗒, 𝖺𝗆𝗈𝗇𝗀 𝗆𝖺𝗇𝗒 𝗍𝗈 𝗆𝗈𝗏𝖾 𝖿𝗈𝗋𝗐𝖺𝗋𝖽 𝗎𝗌𝗂𝗇𝗀 𝗈𝗎𝗋 𝗋𝖾𝗅𝖺𝗍𝗂𝗈𝗇𝗌𝗁𝗂𝗉𝗌. 𝖠𝗇𝖽 𝗍𝗁𝗂𝗌 𝗁𝗈𝗅𝖽𝗌 𝗍𝗁𝖾 𝖼𝗅𝗎𝖾. 𝖶𝗁𝖾𝗇 𝗐𝖾 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗍𝗁𝖾 𝗆𝖾𝗆𝗈 𝗍𝖺𝖻𝗅𝖾 𝖺𝗇𝖽 𝗍𝗁𝖾 𝖺𝖽𝗃𝖺𝖼𝖾𝗇𝗍 𝗉𝖺𝗋𝖾𝗇𝗍 𝖾𝗇𝗍𝗋𝗂𝖾𝗌 𝗂𝗍 𝖻𝖾𝖼𝗈𝗆𝖾𝗌 𝖺𝖻𝗎𝗇𝖽𝖺𝗇𝗍𝗅𝗒 𝖼𝗅𝖾𝖺𝗋 𝗐𝗁𝖾𝗋𝖾 𝗐𝖾 𝖼𝖺𝗆𝖾 𝖿𝗋𝗈𝗆 𝗂𝖿 𝗐𝖾 𝖺𝗇𝖺𝗅𝗒𝗓𝖾 𝖺𝗇𝖽 𝖼𝗈𝗆𝗉𝖺𝗋𝖾 𝗍𝗁𝖾 𝖾𝗇𝗍𝗋𝗂𝖾𝗌 (𝗎𝗌𝗂𝗇𝗀 𝗍𝗁𝖾 𝗋𝖾𝗅𝖺𝗍𝗂𝗈𝗇𝗌𝗁𝗂𝗉 𝖺𝗌 𝗍𝗁𝖾 𝗀𝗎𝗂𝖽𝖾). 𝖨𝗍'𝗌 𝗊𝗎𝗂𝗍𝖾 𝗌𝗍𝗋𝖺𝗂𝗀𝗁𝗍𝖿𝗈𝗋𝗐𝖺𝗋𝖽 𝖻𝗎𝗍 𝗂𝗍 𝖽𝗈𝖾𝗌 𝗁𝖾𝗅𝗉 𝗍𝗈 𝗅𝗈𝗈𝗄 𝖺 𝗏𝖺𝗋𝗂𝖾𝗍𝗒 𝗈𝖿 𝖾𝗑𝖺𝗆𝗉𝗅𝖾𝗌 𝖺𝗇𝖽 𝗌𝗈 𝗒𝗈𝗎 𝖼𝖺𝗇 𝖿𝗂𝗇𝖽 𝗍𝗁𝖾𝗆 𝖻𝖾𝗅𝗈𝗐.

𝖠𝗇𝗈𝗍𝗁𝖾𝗋 𝖾𝗊𝗎𝗂𝗏𝖺𝗅𝖾𝗇𝗍 𝗐𝖺𝗒 𝗍𝗈 𝗌𝗈𝗅𝗏𝖾 𝗍𝗁𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗂𝗌 𝗍𝗈 𝗂𝗆𝖺𝗀𝗂𝗇𝖾 𝗍𝗁𝖾 𝗍𝗁𝖾 𝖿𝗂𝗀𝗎𝗋𝖾 𝗍𝗈 𝖻𝖾 𝖺 𝗗𝗔𝗚 𝖺𝗇𝖽 𝗍𝗁𝖾 𝙚𝙙𝙜𝙚 𝙬𝙚𝙞𝙜𝙝𝙩 𝗈𝖿 𝖺 𝙙𝙞𝙧𝙚𝙘𝙩𝙚𝙙 𝖾𝖽𝗀𝖾 𝘂𝘃 𝗍𝗈 𝖻𝖾

-(𝗟𝗖𝗦(𝘃) - 𝗟𝗖𝗦(𝘂))

𝖳𝗁𝗂𝗌 𝗆𝖺𝗄𝖾𝗌 𝗍𝗁𝖾 𝗀𝗋𝖺𝗉𝗁 𝖾𝖽𝗀𝖾𝗌 𝗆𝗈𝗌𝗍𝗅𝗒 𝟢 𝗂𝗇 𝗐𝖾𝗂𝗀𝗁𝗍 𝗈𝗋 𝗇𝖾𝗀𝖺𝗍𝗂𝗏𝖾. 𝖠𝗇𝖽 𝖿𝗂𝗇𝖽𝗂𝗇𝗀 𝗍𝗁𝖾 𝗦𝗵𝗼𝗿𝘁𝗲𝘀𝘁 𝗣𝗮𝘁𝗵 𝗂𝗇 𝗍𝗁𝗂𝗌 𝗗𝗔𝗚 𝗌𝗍𝖺𝗋𝗍𝗂𝗇𝗀 𝖿𝗋𝗈𝗆

𝘀 = 𝗟[ 𝗺 ][ 𝗻 ] 𝗍𝗈

𝘁 = 𝗟[ 𝟢 ][ 𝟢 ]

𝗐𝗈𝗎𝗅𝖽 𝖺𝗅𝗌𝗈 𝗅𝖾𝖺𝖽 𝗍𝗈 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝗌𝗈𝗅𝗎𝗍𝗂𝗈𝗇. 𝖠𝗀𝖺𝗂𝗇 𝗂𝗍'𝗌 𝗎𝗌𝖾𝖿𝗎𝗅 𝗍𝗈 𝗄𝖾𝖾𝗉 𝗂𝗇 𝗆𝗂𝗇𝖽 𝗍𝗁𝖾 𝗉𝖺𝗍𝗁 𝗂𝗌 𝗎𝗌𝗎𝖺𝗅𝗅𝗒 𝗇𝗈𝗍 𝗎𝗇𝗂𝗊𝗎𝖾 𝖺𝗇𝖽 𝗐𝖾 𝖻𝗋𝖾𝖺𝗄 𝗍𝗂𝖾𝗌 𝖺𝗋𝖻𝗂𝗍𝗋𝖺𝗋𝗂𝗅𝗒 𝖺𝗅𝗈𝗇𝗀 𝗍𝗁𝖾 𝗐𝖺𝗒. 𝖳𝗁𝖾𝗋𝖾 𝖼𝗈𝗎𝗅𝖽 𝖻𝖾 𝗆𝖺𝗇𝗒 𝖫𝖢𝖲𝗌 𝖺𝗇𝖽 𝖾𝖺𝖼𝗁 𝖫𝖢𝖲 𝖼𝗈𝗎𝗅𝖽 𝗉𝗈𝗍𝖾𝗇𝗍𝗂𝖺𝗅𝗅𝗒 𝗁𝖺𝗏𝖾 𝗆𝗎𝗅𝗍𝗂𝗉𝗅𝖾 𝗉𝖺𝗍𝗁𝗌 𝖺𝗌𝗌𝗈𝖼𝗂𝖺𝗍𝖾𝖽 𝗐𝗂𝗍𝗁 𝗂𝗍. 𝖳𝗁𝖾 𝗀𝗈𝖺𝗅 𝗂𝗌 𝗎𝗌𝗎𝖺𝗅𝗅𝗒 𝗍𝗈 𝗈𝗎𝗍𝗉𝗎𝗍 𝗈𝗇𝖾.

Longest Common Subsequence | Tracing Parent Pointers | LCS of THEIR - HABIT is (to say) HI

𝖨𝖨. 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲

𝙇𝘾 𝘔𝘦𝘥𝘪𝘶𝘮 72. 𝙀𝙙𝙞𝙩 𝘿𝙞𝙨𝙩𝙖𝙣𝙘𝙚

𝙈𝙚𝙙𝙞𝙪𝙢

𝘎𝘪𝘷𝘦𝘯 𝘵𝘸𝘰 𝘴𝘵𝘳𝘪𝘯𝘨𝘴 𝘸𝘰𝘳𝘥1 𝘢𝘯𝘥 𝘸𝘰𝘳𝘥2, 𝘳𝘦𝘵𝘶𝘳𝘯 𝘵𝘩𝘦 𝘮𝘪𝘯𝘪𝘮𝘶𝘮 𝘯𝘶𝘮𝘣𝘦𝘳 𝘰𝘧 𝘰𝘱𝘦𝘳𝘢𝘵𝘪𝘰𝘯𝘴 𝘳𝘦𝘲𝘶𝘪𝘳𝘦𝘥 𝘵𝘰 𝘤𝘰𝘯𝘷𝘦𝘳𝘵 𝘸𝘰𝘳𝘥1 𝘵𝘰 𝘸𝘰𝘳𝘥2.

𝘠𝘰𝘶 𝘩𝘢𝘷𝘦 𝘵𝘩𝘦 𝘧𝘰𝘭𝘭𝘰𝘸𝘪𝘯𝘨 𝘵𝘩𝘳𝘦𝘦 𝘰𝘱𝘦𝘳𝘢𝘵𝘪𝘰𝘯𝘴 𝘱𝘦𝘳𝘮𝘪𝘵𝘵𝘦𝘥 𝘰𝘯 𝘢 𝘸𝘰𝘳𝘥:

𝘐𝘯𝘴𝘦𝘳𝘵 𝘢 𝘤𝘩𝘢𝘳𝘢𝘤𝘵𝘦𝘳

𝘋𝘦𝘭𝘦𝘵𝘦 𝘢 𝘤𝘩𝘢𝘳𝘢𝘤𝘵𝘦𝘳

𝘙𝘦𝘱𝘭𝘢𝘤𝘦 𝘢 𝘤𝘩𝘢𝘳𝘢𝘤𝘵𝘦𝘳

𝙀𝙭𝙖𝙢𝙥𝙡𝙚 1:

𝙄𝙣𝙥𝙪𝙩: 𝘸𝘰𝘳𝘥1 = "𝘩𝘰𝘳𝘴𝘦", 𝘸𝘰𝘳𝘥2 = "𝘳𝘰𝘴"

𝙊𝙪𝙩𝙥𝙪𝙩: 3

𝙀𝙭𝙥𝙡𝙖𝙣𝙖𝙩𝙞𝙤𝙣:

𝘩𝘰𝘳𝘴𝘦 -> 𝘳𝘰𝘳𝘴𝘦 (𝘳𝘦𝘱𝘭𝘢𝘤𝘦 '𝘩' 𝘸𝘪𝘵𝘩 '𝘳')

𝘳𝘰𝘳𝘴𝘦 -> 𝘳𝘰𝘴𝘦 (𝘳𝘦𝘮𝘰𝘷𝘦 '𝘳')

𝘳𝘰𝘴𝘦 -> 𝘳𝘰𝘴 (𝘳𝘦𝘮𝘰𝘷𝘦 '𝘦')

𝙀𝙭𝙖𝙢𝙥𝙡𝙚 2:

𝙄𝙣𝙥𝙪𝙩: 𝘸𝘰𝘳𝘥1 = "𝘪𝘯𝘵𝘦𝘯𝘵𝘪𝘰𝘯", 𝘸𝘰𝘳𝘥2 = "𝘦𝘹𝘦𝘤𝘶𝘵𝘪𝘰𝘯"

𝙊𝙪𝙩𝙥𝙪𝙩: 5

𝙀𝙭𝙥𝙡𝙖𝙣𝙖𝙩𝙞𝙤𝙣:

𝘪𝘯𝘵𝘦𝘯𝘵𝘪𝘰𝘯 -> 𝘪𝘯𝘦𝘯𝘵𝘪𝘰𝘯 (𝘳𝘦𝘮𝘰𝘷𝘦 '𝘵')

𝘪𝘯𝘦𝘯𝘵𝘪𝘰𝘯 -> 𝘦𝘯𝘦𝘯𝘵𝘪𝘰𝘯 (𝘳𝘦𝘱𝘭𝘢𝘤𝘦 '𝘪' 𝘸𝘪𝘵𝘩 '𝘦')

𝘦𝘯𝘦𝘯𝘵𝘪𝘰𝘯 -> 𝘦𝘹𝘦𝘯𝘵𝘪𝘰𝘯 (𝘳𝘦𝘱𝘭𝘢𝘤𝘦 '𝘯' 𝘸𝘪𝘵𝘩 '𝘹')

𝘦𝘹𝘦𝘯𝘵𝘪𝘰𝘯 -> 𝘦𝘹𝘦𝘤𝘵𝘪𝘰𝘯 (𝘳𝘦𝘱𝘭𝘢𝘤𝘦 '𝘯' 𝘸𝘪𝘵𝘩 '𝘤')

𝘦𝘹𝘦𝘤𝘵𝘪𝘰𝘯 -> 𝘦𝘹𝘦𝘤𝘶𝘵𝘪𝘰𝘯 (𝘪𝘯𝘴𝘦𝘳𝘵 '𝘶')

𝟣. 𝗦𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀

𝖩𝗎𝗌𝗍 𝗅𝗂𝗄𝖾 𝗐𝗂𝗍𝗁 𝗍𝗁𝖾 𝗉𝗋𝖾𝗏𝗂𝗈𝗎𝗌 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗟𝗖𝗦 𝗐𝖾 𝗀𝗈 𝗐𝗂𝗍𝗁 𝗍𝗁𝖾 𝗣𝗿𝗼𝗱𝘂𝗰𝘁 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗦𝘂𝗳𝗳𝗶𝘅𝗲𝘀:

𝗗[ 𝗶 ][ 𝗷 ] 𝗍𝗁𝖾 𝗆𝗂𝗇𝗂𝗆𝗎𝗆 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝖻𝖾𝗍𝗐𝖾𝖾𝗇 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗌𝗎𝖿𝖿𝗂𝗑 𝗔[𝗶 : ] 𝖺𝗇𝖽 𝗕[𝗷 : ]

𝗗[ 𝗶 ][ 𝗷 ] = 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 (𝗔[𝗶 : ], 𝗕[𝗷 : ])

𝖿𝗈𝗋

𝟢 <= 𝗶 <= |𝗔| (𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝗔 = 𝗺) 𝖺𝗇𝖽

𝟢 <= 𝗷 <= |𝗕| (𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝗕 = 𝗻)

Edit Distance | Subproblems | Suffixes of A X Suffixes of B
Edit Distance | Relationships | Alignment Possibilities | A[i] and B[j] Match
Edit Distance | Relationships | Alignment Possibilities | A[i] and B[j] Mismatch
Edit Distance | Relationships | Alignment Possibilities | A[i] is an Insertion


Edit Distance | Relationships | Alignment Possibilities | A[i] is a Deletion
Edit Distance | Subproblem Relationships | Putting it all together

𝟤. 𝗥𝗲𝗹𝗮𝘁𝗶𝗼𝗻𝘀𝗵𝗶𝗽𝘀

𝖠𝗌 𝗒𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝗂𝗇 𝗍𝗁𝖾 𝖿𝗂𝗀𝗎𝗋𝖾𝗌 𝖺𝖻𝗈𝗏𝖾:

𝖺. 𝖶𝗁𝖾𝗇 𝗍𝗁𝖾𝗋𝖾'𝗌 𝖺 𝗺𝗮𝘁𝗰𝗵 𝗈𝖿 𝗔[ 𝗶 ] 𝖺𝗇𝖽 𝗕[ 𝗷 ]: 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗗[ 𝗶 ][ 𝗷 ] 𝗂𝗌 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝖺𝗌 𝗗[𝗶 + 𝟣][𝗷 + 𝟣]

𝖻. 𝖶𝗁𝖾𝗇 𝗍𝗁𝖾𝗋𝖾'𝗌 𝖺 𝗺𝗶𝘀𝗺𝗮𝘁𝗰𝗵 𝗈𝖿 𝗔[ 𝗶 ] 𝖺𝗇𝖽 𝗕[ 𝗷 ]: 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗗[ 𝗶 ][ 𝗷 ] 𝗂𝗌 𝗗[𝗶 + 𝟣][𝗷 + 𝟣] + 𝟣 (𝖺𝗌 𝗐𝖾 𝗐𝗈𝗎𝗅𝖽 𝗇𝖾𝖾𝖽 𝗍𝗈 𝗋𝖾𝗉𝗅𝖺𝖼𝖾 𝗍𝗁𝖾 𝗆𝗂𝗌𝗆𝖺𝗍𝖼𝗁𝗂𝗇𝗀 𝖼𝗁𝖺𝗋 𝗅𝖾𝖺𝖽𝗂𝗇𝗀 𝗍𝗈 𝖺 𝗌𝗂𝗇𝗀𝗅𝖾 𝗈𝗉𝖾𝗋𝖺𝗍𝗂𝗈𝗇)

𝖼. 𝖶𝗁𝖾𝗇 𝗔[ 𝗶 ] 𝗂𝗌 𝖺𝗇 𝗜𝗻𝘀𝗲𝗿𝘁𝗶𝗼𝗻: 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗗[ 𝗶 ][ 𝗷 ] 𝗂𝗌 𝗗[𝗂 + 𝟣][ 𝗷 ] + 𝟣 (𝖺𝖼𝖼𝗈𝗎𝗇𝗍𝗂𝗇𝗀 𝖿𝗈𝗋 𝟣 𝗜𝗻𝘀𝗲𝗿𝘁)

𝖽. 𝖶𝗁𝖾𝗇 𝗔[ 𝗶 ] 𝗂𝗌 𝖺 𝗗𝗲𝗹𝗲𝘁𝗶𝗼𝗻: 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗗[ 𝗶 ][ 𝗷 ] 𝗂𝗌 𝗗[ 𝗶 ][𝗷 + 𝟣] + 𝟣 (𝖺𝖼𝖼𝗈𝗎𝗇𝗍𝗂𝗇𝗀 𝖿𝗈𝗋 𝟣 𝗗𝗲𝗹𝗲𝘁𝗲)

𝖠𝗇𝖽 𝗍𝗁𝗂𝗌 𝗂𝗌 𝗐𝗁𝖺𝗍 𝗂𝗍 𝗅𝗈𝗈𝗄𝗌 𝗅𝗂𝗄𝖾 𝗂𝗇 𝖼𝗈𝖽𝖾:

Edit Distance | Subproblem Relationships

𝟥. 𝗧𝗼𝗽𝗼𝗹𝗼𝗴𝗶𝗰𝗮𝗹 𝗢𝗿𝗱𝗲𝗿

𝖲𝗂𝗇𝖼𝖾 𝗐𝖾 𝖽𝗈𝗇'𝗍 𝗐𝖺𝗇𝗍 𝗍𝗈 𝗂𝗇𝖼𝗎𝗋 𝗍𝗁𝖾 𝖼𝗈𝗌𝗍 𝗈𝖿 𝗍𝗈𝗉 𝖽𝗈𝗐𝗇 𝗋𝖾𝖼𝗎𝗋𝗌𝗂𝗈𝗇 𝗐𝖾 𝗌𝗈𝗅𝗏𝖾 𝖿𝗈𝗋 𝗌𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗂𝗇 𝗍𝗁𝖾 𝖻𝗈𝗍𝗍𝗈𝗆 𝗎𝗉 𝖽𝗂𝗋𝖾𝖼𝗍𝗂𝗈𝗇 𝖺𝗄𝖺 𝗂𝗇𝖼𝗋𝖾𝖺𝗌𝗂𝗇𝗀 𝗌𝗂𝗓𝖾 𝗈𝖿 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗂.𝖾.,

𝗶 𝗀𝗈𝗂𝗇𝗀 𝖿𝗋𝗈𝗆 [𝗺 -> 𝟢] 𝖺𝗇𝖽

𝗷 𝗀𝗈𝗂𝗇𝗀 𝖿𝗋𝗈𝗆 [𝗻 -> 𝟢]

𝟦. 𝗕𝗮𝘀𝗲 𝗖𝗮𝘀𝗲(𝗌)

𝖳𝗁𝖾 𝖡𝖺𝗌𝖾 𝖢𝖺𝗌𝖾 𝗂𝗌 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝗐𝗁𝖾𝗇 𝖾𝗂𝗍𝗁𝖾𝗋 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝖺𝗋𝖾 𝖾𝗆𝗉𝗍𝗒 𝖻𝗎𝗍 𝗂𝗇 𝗍𝗁𝗂𝗌 𝖼𝖺𝗌𝖾 𝗍𝗁𝖾 𝖾𝖽𝗂𝗍 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗈𝖿 𝖺 𝗇𝗈𝗇-𝖾𝗆𝗉𝗍𝗒 𝗌𝗍𝗋𝗂𝗇𝗀 𝗍𝗈 𝖺𝗇 𝖾𝗆𝗉𝗍𝗒 𝗌𝗍𝗋𝗂𝗇𝗀 / 𝗌𝗎𝖿𝖿𝗂𝗑 𝗂𝗌 𝗍𝗁𝖾 𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝗍𝗁𝖾 𝗇𝗈𝗇-𝖾𝗆𝗉𝗍𝗒 𝗌𝗍𝗋𝗂𝗇𝗀 𝖻𝖾𝖼𝖺𝗎𝗌𝖾 𝗐𝖾 𝗁𝖺𝗏𝖾 𝗍𝗈 𝗂𝗇𝗌𝖾𝗋𝗍 𝖺𝗅𝗅 𝗍𝗁𝖾 𝖼𝗁𝖺𝗋𝖺𝖼𝗍𝖾𝗋𝗌 𝗍𝗈 𝗍𝗋𝖺𝗇𝗌𝖿𝗈𝗋𝗆 𝗍𝗁𝖾 𝖾𝗆𝗉𝗍𝗒 𝗌𝗍𝗋𝗂𝗇𝗀. 𝖠𝗅𝗌𝗈 𝗄𝖾𝖾𝗉 𝗂𝗇 𝗆𝗂𝗇𝖽 𝗍𝗁𝖺𝗍 𝗍𝗁𝖾

𝖺. 𝗟𝗲𝗻𝗴𝘁𝗵 𝗈𝖿 𝗌𝗎𝖿𝖿𝗂𝗑 𝗔[𝗶 : ] = 𝗺 - 𝗶 𝖺𝗇𝖽 𝗍𝗁𝖾

𝖻. 𝗟𝗲𝗻𝗴𝘁𝗵 𝗈𝖿 𝗌𝗎𝖿𝖿𝗂𝗑 𝗕[𝗷 : ] = 𝗻 - 𝗷

𝗍𝗁𝖾𝗌𝖾 𝖺𝗋𝖾 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝖺𝗇𝖽 𝗇𝗈𝗍 𝗉𝗋𝖾𝖿𝗂𝗑𝖾𝗌 𝗌𝗈 𝗐𝖾 𝗇𝖾𝖾𝖽 𝗍𝗈 𝗄𝖾𝖾𝗉 𝗍𝗁𝖺𝗍 𝗂𝗇 𝗆𝗂𝗇𝖽. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗈𝗎𝗋 𝖻𝖺𝗌𝖾 𝖼𝖺𝗌𝖾𝗌 𝖺𝗋𝖾:

Edit Distance | Base Cases | When either of the suffix is empty the edit distance of the non-empty string from / to the empty suffix is just the length of the suffix which is n - j and m - i for B[j : ] and A[i : ]

𝟧. 𝗢𝗿𝗶𝗴𝗶𝗻𝗮𝗹 𝗣𝗿𝗼𝗯𝗹𝗲𝗺

𝖳𝗁𝖾 𝗈𝗋𝗂𝗀𝗂𝗇𝖺𝗅 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝗋𝖾𝗊𝗎𝗂𝗋𝖾𝗌 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝖿𝗈𝗋 𝗍𝗁𝖾 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 𝖻𝖾𝗍𝗐𝖾𝖾𝗇 𝖻𝗈𝗍𝗁 𝗈𝖿 𝗍𝗁𝖾 𝖾𝗇𝗍𝗂𝗋𝖾 𝗌𝗍𝗋𝗂𝗇𝗀𝗌 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗗[ 𝟢 ][ 𝟢 ]

𝟨. 𝗧𝗶𝗺𝗲 𝖢𝗈𝗆𝗉𝗅𝖾𝗑𝗂𝗍𝗒

𝖲𝗂𝗇𝖼𝖾 𝗐𝖾 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋 𝖺𝗅𝗅 𝗉𝖺𝗂𝗋𝗌 𝗈𝖿 𝗂𝗇𝖽𝖾𝗉𝖾𝗇𝖽𝖾𝗇𝗍 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌 𝗍𝗁𝖾

𝗻𝘂𝗺𝗯𝗲𝗿 𝗈𝖿 𝘀𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀 ~ 𝗢(𝗻^𝟤)

𝘁𝗶𝗺𝗲 𝗉𝖾𝗋 𝘀𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺 𝗂𝗌 𝖼𝗅𝖾𝖺𝗋𝗅𝗒 𝖼𝗈𝗇𝗌𝗍𝖺𝗇𝗍 𝗍𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗢(𝟣)

𝗍𝗁𝖾 𝗈𝗏𝖾𝗋𝖺𝗅𝗅 𝗍𝗂𝗆𝖾 𝖼𝗈𝗆𝗉𝗅𝖾𝗑𝗂𝗍𝗒 𝗍𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝗂𝗌 𝗢(𝗻^𝟤)

𝖫𝖾𝗍'𝗌 𝗌𝖾𝖾 𝗐𝗁𝖺𝗍 𝗍𝗁𝗂𝗌 𝗅𝗈𝗈𝗄𝗌 𝗅𝗂𝗄𝖾 𝗂𝗇 𝖼𝗈𝖽𝖾:

Edit Distance | Solution

𝖭𝗈𝗍𝗂𝖼𝖾 𝗁𝗈𝗐 𝗂𝗇 𝗍𝗁𝗂𝗌 𝖼𝖺𝗌𝖾 𝗐𝖾 𝖺𝗋𝖾 𝗂𝗇𝗂𝗍𝗂𝖺𝗅𝗂𝗓𝗂𝗇𝗀 𝖺𝗅𝗅 𝗏𝖺𝗅𝗎𝖾𝗌 𝖾𝗑𝖼𝖾𝗉𝗍 𝗍𝗁𝖾 𝖻𝖺𝗌𝖾 𝖼𝖺𝗌𝖾𝗌 𝗍𝗈 𝖻𝖾 𝗶𝗻𝗳 𝖺𝗄𝖺 𝗳𝗹𝗼𝗮𝘁('𝗶𝗻𝗳'). 𝖳𝗁𝖾 𝗋𝖾𝖺𝗌𝗈𝗇 𝖻𝖾𝗂𝗇𝗀 𝗍𝗁𝖾𝗋𝖾'𝗌 𝗋𝖾𝖺𝗅𝗅𝗒 𝗇𝗈 𝗎𝗉𝗉𝖾𝗋 𝗅𝗂𝗆𝗂𝗍 𝗈𝗇 𝗍𝗁𝖾 𝗉𝗈𝗌𝗌𝗂𝖻𝗅𝖾 𝗏𝖺𝗅𝗎𝖾 𝗈𝖿 𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝖺 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖺𝗇𝖽 𝗌𝗈 𝗐𝖾 𝗁𝖺𝗏𝖾 𝗍𝗈 𝗉𝗂𝖼𝗄 𝗍𝗁𝖾 𝗺𝗮𝘅 𝗏𝖺𝗅𝗎𝖾 𝗽𝗼𝘀𝘀𝗶𝗯𝗹𝗲.

𝟩. 𝗕 (𝖼𝖺𝗇 𝖺𝗅𝗌𝗈 𝖻𝖾) 𝖿𝗈𝗋 𝗕𝗮𝗰𝗸𝘁𝗿𝗮𝗰𝗸𝗶𝗻𝗴 (𝖳𝗋𝖺𝖼𝗂𝗇𝗀) 𝖯𝖺𝗋𝖾𝗇𝗍 𝗣𝗼𝗶𝗻𝘁𝗲𝗿𝘀

𝖠𝗌 𝗐𝖾 𝗁𝖺𝗏𝖾 𝖺𝗅𝗋𝖾𝖺𝖽𝗒 𝗆𝖾𝗇𝗍𝗂𝗈𝗇𝖾𝖽 𝗍𝗁𝖺𝗍 𝗆𝗈𝗌𝗍 𝗗𝘆𝗻𝗮𝗺𝗶𝗰 𝗣𝗿𝗼𝗴𝗿𝗮𝗺𝗺𝗶𝗻𝗴 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝖼𝖺𝗇 𝖻𝖾 𝗿𝗲𝗱𝘂𝗰𝗲𝗱 𝗍𝗈 𝗌𝗈𝗅𝗏𝗂𝗇𝗀 𝗦𝗵𝗼𝗿𝘁𝗲𝘀𝘁 𝗣𝗮𝘁𝗵𝗌 𝗂𝗇 𝖺 𝗗𝗶𝗿𝗲𝗰𝘁𝗲𝗱 𝗔𝗰𝘆𝗰𝗹𝗶𝗰 𝗚𝗿𝗮𝗽𝗵 (𝖣𝖠𝖦) 𝗈𝖿 𝗦𝘂𝗯𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀. 𝖨𝗇 𝖺 𝗅𝖺𝗋𝗀𝖾 𝗌𝗎𝖻𝗌𝖾𝗍 𝗈𝖿 𝗍𝗁𝖾𝗌𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆𝗌 𝗐𝖾 𝗆𝗂𝗀𝗁𝗍 𝗇𝗈𝗍 𝗃𝗎𝗌𝗍 𝖻𝖾 𝗂𝗇𝗍𝖾𝗋𝖾𝗌𝗍𝖾𝖽 𝗂𝗇 𝖼𝗈𝗆𝗉𝗎𝗍𝗂𝗇𝗀 𝗍𝗁𝖾 𝗉𝗋𝗈𝗉𝖾𝗋𝗍𝗒 𝗈𝖿 𝖺 𝖲𝗁𝗈𝗋𝗍𝖾𝗌𝗍 𝖯𝖺𝗍𝗁 (𝗅𝗂𝗄𝖾 𝗂𝗍𝗌 𝗅𝖾𝗇𝗀𝗍𝗁) 𝗂𝗇 𝖺 𝖣𝖠𝖦 𝖻𝗎𝗍 𝖺𝗅𝗌𝗈 𝗍𝗁𝖾 𝖯𝖺𝗍𝗁 𝗂𝗍𝗌𝖾𝗅𝖿. 𝖠𝗇𝖽 𝖺 𝖼𝗈𝗆𝗉𝗅𝖾𝗍𝖾 𝗆𝖾𝗆𝗈 𝗍𝖺𝖻𝗅𝖾 𝗂𝖿 𝗆𝖺𝗂𝗇𝗍𝖺𝗂𝗇𝖾𝖽 𝗁𝖺𝗌 𝖾𝗇𝗈𝗎𝗀𝗁 𝗂𝗇𝖿𝗈𝗋𝗆𝖺𝗍𝗂𝗈𝗇 𝗍𝗈 𝗌𝗇𝗂𝖿𝖿 𝗍𝗁𝗂𝗌 𝗈𝗎𝗍 𝗈𝖿 𝗂𝗍𝗌𝖾𝗅𝖿. 𝖫𝖾𝗍'𝗌 𝗌𝖾𝖾 𝗁𝗈𝗐 𝗐𝖾 𝖼𝖺𝗇 𝗀𝗈 𝖺𝖻𝗈𝗎𝗍 𝖽𝗈𝗂𝗇𝗀 𝗍𝗁𝖺𝗍 𝗂𝗇 𝗍𝗁𝗂𝗌 𝖲𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖠𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖺𝗄𝖺 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲.

𝖳𝗁𝖾 𝗄𝖾𝗒 𝗍𝗈 𝗎𝗇𝖽𝖾𝗋𝗌𝗍𝖺𝗇𝖽𝗂𝗇𝗀 𝗍𝗋𝖺𝖼𝗂𝗇𝗀 𝗈𝖿 𝗉𝖺𝗋𝖾𝗇𝗍 𝗉𝗈𝗂𝗇𝗍𝖾𝗋𝗌 𝗂𝗌 𝗂𝗇 𝗋𝖾𝖺𝗅𝗂𝗓𝗂𝗇𝗀 𝗍𝗁𝖺𝗍 𝖺𝗍 𝖾𝖺𝖼𝗁 𝗌𝗍𝖾𝗉 𝗐𝖾 𝖼𝗁𝗈𝗈𝗌𝖾 𝖺 𝗉𝗈𝗌𝗌𝗂𝖻𝗂𝗅𝗂𝗍𝗒, 𝖻𝗋𝖾𝖺𝗄𝗂𝗇𝗀 𝗍𝗂𝖾𝗌, 𝗂𝖿 𝗍𝗁𝖾𝗒 𝖾𝗑𝗂𝗌𝗍, 𝖺𝗋𝖻𝗂𝗍𝗋𝖺𝗋𝗂𝗅𝗒, 𝖺𝗆𝗈𝗇𝗀 𝗆𝖺𝗇𝗒 𝗍𝗈 𝗆𝗈𝗏𝖾 𝖿𝗈𝗋𝗐𝖺𝗋𝖽 𝗎𝗌𝗂𝗇𝗀 𝗈𝗎𝗋 𝗋𝖾𝗅𝖺𝗍𝗂𝗈𝗇𝗌𝗁𝗂𝗉𝗌. 𝖠𝗇𝖽 𝗍𝗁𝗂𝗌 𝗁𝗈𝗅𝖽𝗌 𝗍𝗁𝖾 𝖼𝗅𝗎𝖾. 𝖶𝗁𝖾𝗇 𝗐𝖾 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗍𝗁𝖾 𝗆𝖾𝗆𝗈 𝗍𝖺𝖻𝗅𝖾 𝖺𝗇𝖽 𝗍𝗁𝖾 𝖺𝖽𝗃𝖺𝖼𝖾𝗇𝗍 𝗉𝖺𝗋𝖾𝗇𝗍 𝖾𝗇𝗍𝗋𝗂𝖾𝗌 𝗂𝗍 𝖻𝖾𝖼𝗈𝗆𝖾𝗌 𝖺𝖻𝗎𝗇𝖽𝖺𝗇𝗍𝗅𝗒 𝖼𝗅𝖾𝖺𝗋 𝗐𝗁𝖾𝗋𝖾 𝗐𝖾 𝖼𝖺𝗆𝖾 𝖿𝗋𝗈𝗆 𝗂𝖿 𝗐𝖾 𝖺𝗇𝖺𝗅𝗒𝗓𝖾 𝖺𝗇𝖽 𝖼𝗈𝗆𝗉𝖺𝗋𝖾 𝗍𝗁𝖾 𝖾𝗇𝗍𝗋𝗂𝖾𝗌 (𝗎𝗌𝗂𝗇𝗀 𝗍𝗁𝖾 𝗋𝖾𝗅𝖺𝗍𝗂𝗈𝗇𝗌𝗁𝗂𝗉 𝖺𝗌 𝗍𝗁𝖾 𝗀𝗎𝗂𝖽𝖾). 𝖨𝗍'𝗌 𝗊𝗎𝗂𝗍𝖾 𝗌𝗍𝗋𝖺𝗂𝗀𝗁𝗍𝖿𝗈𝗋𝗐𝖺𝗋𝖽 𝖻𝗎𝗍 𝗂𝗍 𝖽𝗈𝖾𝗌 𝗁𝖾𝗅𝗉 𝗍𝗈 𝗅𝗈𝗈𝗄 𝖺 𝗏𝖺𝗋𝗂𝖾𝗍𝗒 𝗈𝖿 𝖾𝗑𝖺𝗆𝗉𝗅𝖾𝗌 𝖺𝗇𝖽 𝗌𝗈 𝗒𝗈𝗎 𝖼𝖺𝗇 𝖿𝗂𝗇𝖽 𝗍𝗁𝖾𝗆 𝖻𝖾𝗅𝗈𝗐.

𝖠𝗇𝗈𝗍𝗁𝖾𝗋 𝖾𝗊𝗎𝗂𝗏𝖺𝗅𝖾𝗇𝗍 𝗐𝖺𝗒 𝗍𝗈 𝗌𝗈𝗅𝗏𝖾 𝗍𝗁𝖾 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗂𝗌 𝗍𝗈 𝗂𝗆𝖺𝗀𝗂𝗇𝖾 𝗍𝗁𝖾 𝗍𝗁𝖾 𝖿𝗂𝗀𝗎𝗋𝖾 𝗍𝗈 𝖻𝖾 𝖺 𝗗𝗔𝗚 𝖺𝗇𝖽 𝗍𝗁𝖾 𝙚𝙙𝙜𝙚 𝙬𝙚𝙞𝙜𝙝𝙩 𝗈𝖿 𝖺 𝙙𝙞𝙧𝙚𝙘𝙩𝙚𝙙 𝖾𝖽𝗀𝖾 𝘂𝘃 𝗍𝗈 𝖻𝖾

𝗗(𝘃) - 𝗗(𝘂)

(𝗡𝗼𝘁𝗲: 𝗍𝗁𝖾 𝗆𝗂𝗌𝗌𝗂𝗇𝗀 𝗇𝖾𝗀𝖺𝗍𝗂𝗏𝖾 𝗌𝗂𝗀𝗇 𝗂𝗇 𝗍𝗁𝗂𝗌 𝖼𝖺𝗌𝖾 𝖺𝗇𝖽 𝗍𝗁𝖺𝗍'𝗌 𝖻𝖾𝖼𝖺𝗎𝗌𝖾 𝗐𝖾 𝖺𝗋𝖾 𝗅𝗈𝗈𝗄𝗂𝗇𝗀 𝗍𝗈 𝗆𝗂𝗇𝗂𝗆𝗂𝗓𝖾 𝗍𝗁𝖾 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 𝗎𝗇𝗅𝗂𝗄𝖾 𝗂𝗇 𝗍𝗁𝖾 𝖼𝖺𝗌𝖾 𝗈𝖿 𝗟𝗖𝗦 𝗐𝗁𝖾𝗋𝖾 𝗐𝖾 𝗐𝖾𝗋𝖾 𝖺𝗂𝗆𝗂𝗇𝗀 𝗍𝗈 𝗆𝖺𝗑𝗂𝗆𝗂𝗓𝖾)

𝖠𝗇𝖽 𝖿𝗂𝗇𝖽𝗂𝗇𝗀 𝗍𝗁𝖾 𝗦𝗵𝗼𝗿𝘁𝗲𝘀𝘁 𝗣𝗮𝘁𝗵 𝗂𝗇 𝗍𝗁𝗂𝗌 𝗗𝗔𝗚 𝗌𝗍𝖺𝗋𝗍𝗂𝗇𝗀 𝖿𝗋𝗈𝗆

𝘀 = 𝗗[ 𝟢 ][ 𝟢 ] 𝗍𝗈

𝘁 = 𝗗[ 𝗺 ][ 𝗻 ]

(𝗡𝗼𝘁𝗲: 𝗍𝗁𝖾 𝗈𝗋𝖽𝖾𝗋 𝗂𝗌 𝗋𝖾𝗏𝖾𝗋𝗌𝖾𝖽 𝖻𝖾𝖼𝖺𝗎𝗌𝖾 𝗈𝗎𝗋 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝖣𝖠𝖦𝗌 𝖺𝗇𝖽 𝗆𝖾𝗆𝗈 𝗍𝖺𝖻𝗅𝖾𝗌 𝖻𝖾𝗅𝗈𝗐 𝗎𝗌𝖾 𝗉𝗋𝖾𝖿𝗂𝗑𝖾𝗌 𝖺𝗇𝖽 𝗇𝗈𝗍 𝗌𝗎𝖿𝖿𝗂𝗑𝖾𝗌)

𝗐𝗈𝗎𝗅𝖽 𝖺𝗅𝗌𝗈 𝗅𝖾𝖺𝖽 𝗍𝗈 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝗌𝗈𝗅𝗎𝗍𝗂𝗈𝗇. 𝖠𝗀𝖺𝗂𝗇 𝗂𝗍'𝗌 𝗎𝗌𝖾𝖿𝗎𝗅 𝗍𝗈 𝗄𝖾𝖾𝗉 𝗂𝗇 𝗆𝗂𝗇𝖽 𝗍𝗁𝖾 𝗉𝖺𝗍𝗁 𝗂𝗌 𝗎𝗌𝗎𝖺𝗅𝗅𝗒 𝗇𝗈𝗍 𝗎𝗇𝗂𝗊𝗎𝖾 𝖺𝗇𝖽 𝗐𝖾 𝖻𝗋𝖾𝖺𝗄 𝗍𝗂𝖾𝗌 𝖺𝗋𝖻𝗂𝗍𝗋𝖺𝗋𝗂𝗅𝗒 𝖺𝗅𝗈𝗇𝗀 𝗍𝗁𝖾 𝗐𝖺𝗒. 𝖳𝗁𝖾𝗋𝖾 𝖼𝗈𝗎𝗅𝖽 𝖻𝖾 𝗆𝖺𝗇𝗒 𝗢𝗽𝘁𝗶𝗺𝗮𝗹 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁𝗌 𝖺𝗇𝖽 𝖾𝖺𝖼𝗁 𝗢𝗽𝘁𝗶𝗺𝗮𝗹 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁 𝖼𝗈𝗎𝗅𝖽 𝗉𝗈𝗍𝖾𝗇𝗍𝗂𝖺𝗅𝗅𝗒 𝗁𝖺𝗏𝖾 𝗆𝗎𝗅𝗍𝗂𝗉𝗅𝖾 𝗉𝖺𝗍𝗁𝗌 (𝗦𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝗈𝖿 𝗘𝗱𝗶𝘁 𝗢𝗽𝗲𝗿𝗮𝘁𝗶𝗼𝗻𝗌) 𝖺𝗌𝗌𝗈𝖼𝗂𝖺𝗍𝖾𝖽 𝗐𝗂𝗍𝗁 𝗂𝗍. 𝖳𝗁𝖾 𝗀𝗈𝖺𝗅 𝗂𝗌 𝗎𝗌𝗎𝖺𝗅𝗅𝗒 𝗍𝗈 𝗈𝗎𝗍𝗉𝗎𝗍 𝗈𝗇𝖾.

Edit Distance | Tracing Parent Pointers | Base Cases

𝖭𝗈𝗍𝗂𝖼𝖾 𝗍𝗁𝖺𝗍 𝗍𝗁𝖾 𝗿𝗲𝗱 𝗮𝗿𝗿𝗼𝘄𝗌 𝗍𝗈 𝗍𝗁𝖾 𝗿𝗶𝗴𝗵𝘁 𝖼𝗈𝗋𝗋𝖾𝗌𝗉𝗈𝗇𝖽 𝗍𝗈 𝗗𝗲𝗹𝗲𝘁𝗶𝗼𝗻𝗌. 𝖳𝗁𝖾 𝖾𝖺𝗌𝗒 𝗐𝖺𝗒 𝗍𝗈 𝖿𝗂𝗀𝗎𝗋𝖾 𝗍𝗁𝗂𝗌 𝗈𝗎𝗍 𝗂𝗌 𝗍𝗈 𝗌𝖾𝖾 𝗍𝗁𝖺𝗍 𝗐𝖾 𝖺𝗋𝖾 𝗍𝗋𝗒𝗂𝗇𝗀 𝗍𝗈 𝗍𝗋𝖺𝗇𝗌𝖿𝗈𝗋𝗆 𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝗍𝗈 𝗘𝗗𝗜𝗧𝗜𝗡𝗚. 𝖲𝗈 𝗐𝗁𝖾𝗇 𝗒𝗈𝗎 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗍𝗁𝖾 𝗉𝖺𝗍𝗁 𝖿𝗋𝗈𝗆

𝗗[ 𝟢 ][ 𝟢 ] 𝗍𝗈

𝗗[ 𝟢 ][ 𝟪 ]

𝗐𝗁𝖺𝗍 𝗍𝗁𝖾 𝘇𝗲𝗿𝗼 𝗈𝗇 𝗍𝗁𝖾 𝗶 𝗂𝗇𝖽𝖾𝗑 (𝗘𝗗𝗜𝗧𝗜𝗡𝗚 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾) 𝗆𝖾𝖺𝗇𝗌 𝗂𝗌 𝗍𝗁𝖺𝗍 𝗐𝖾 𝖺𝗋𝖾 𝗅𝗈𝗈𝗄𝗂𝗇𝗀 𝖺𝗍 𝗍𝗁𝖾 𝗲𝗺𝗽𝘁𝘆 𝗣𝗿𝗲𝗳𝗶𝘅 𝗈𝖿 𝗘𝗗𝗜𝗧𝗜𝗡𝗚. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾 𝖺𝗇𝗒 𝗅𝖾𝗇𝗀𝗍𝗁 𝗈𝖿 𝗍𝗁𝖾 𝗣𝗿𝗲𝗳𝗶𝘅 𝗈𝖿 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝖼𝖺𝗇 𝗈𝗇𝗅𝗒 𝖻𝖾 𝗍𝗋𝖺𝗇𝗌𝖿𝗈𝗋𝗆𝖾𝖽 𝗍𝗁𝗋𝗈𝗎𝗀𝗁 𝗗𝗲𝗹𝗲𝘁𝗶𝗼𝗻𝗌 𝗈𝖿 𝖺𝗅𝗅 𝖼𝗁𝖺𝗋𝖺𝖼𝗍𝖾𝗋𝗌. 𝖲𝗂𝗆𝗂𝗅𝖺𝗋𝗅𝗒 𝗍𝗁𝖾 𝗀𝗋𝖾𝖾𝗇 𝖺𝗋𝗋𝗈𝗐𝗌 𝗍𝗈𝗐𝖺𝗋𝖽𝗌 𝗍𝗁𝖾 𝖻𝗈𝗍𝗍𝗈𝗆 𝖼𝗈𝗋𝗋𝖾𝗌𝗉𝗈𝗇𝖽 𝗍𝗈 𝗜𝗻𝘀𝗲𝗿𝘁𝗶𝗼𝗻𝗌. 𝖬𝖺𝗄𝖾 𝗌𝗎𝗋𝖾 𝗒𝗈𝗎 𝖼𝗈𝗇𝗏𝗂𝗇𝖼𝖾 𝗒𝗈𝗎𝗋𝗌𝖾𝗅𝖿 𝗈𝖿 𝗍𝗁𝗂𝗌.

Edit Distance | Small Example Path | 2 Different Paths from D[0][0] to D[2][2]

𝖭𝗈𝗐 𝗐𝖾 𝗅𝗈𝗈𝗄 𝖺𝗍 𝖺 𝗌𝗅𝗂𝗀𝗁𝗍𝗅𝗒 𝗆𝗈𝗋𝖾 𝗂𝗇𝗏𝗈𝗅𝗏𝖾𝖽 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝗈𝖿 𝗍𝗋𝖺𝗏𝖾𝗋𝗌𝗂𝗇𝗀 𝖿𝗋𝗈𝗆

𝗗[ 𝟢 ][ 𝟢 ] 𝗍𝗈

𝗗[ 𝟤 ][ 𝟤 ]

𝖠𝗌 𝗒𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝗍𝗁𝖾𝗋𝖾 𝖺𝗋𝖾 𝘁𝘄𝗼 𝗉𝗈𝗌𝗌𝗂𝖻𝗅𝖾 𝗽𝗮𝘁𝗵𝘀 𝗆𝖺𝗋𝗄𝖾𝖽 𝗐𝗂𝗍𝗁 𝗱𝗼𝘁𝘁𝗲𝗱 𝖻𝗅𝖺𝖼𝗄 𝖺𝗋𝗋𝗈𝗐𝗌. 𝖠𝗇 𝗮𝗿𝗿𝗼𝘄 𝗍𝗈 𝗍𝗁𝖾 𝗿𝗶𝗴𝗵𝘁 𝗂𝗌 𝖺 𝗗𝗲𝗹𝗲𝘁𝗶𝗼𝗻, 𝖺𝗇 𝗮𝗿𝗿𝗼𝘄 𝗍𝗈 𝗍𝗁𝖾 𝖻𝗈𝗍𝗍𝗈𝗆 𝗂𝗌 𝖺𝗇 𝗜𝗻𝘀𝗲𝗿𝘁𝗶𝗼𝗻. 𝖠 𝗱𝗼𝘄𝗻𝘄𝗮𝗿𝗱 𝗱𝗶𝗮𝗴𝗼𝗻𝗮𝗹 𝗮𝗿𝗿𝗼𝘄 𝖼𝖺𝗇 𝖻𝖾 𝖾𝗂𝗍𝗁𝖾𝗋 𝖺 𝗠𝗮𝘁𝗰𝗵 𝗈𝗋 𝖺 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵. 𝖳𝗁𝖾 𝗐𝖺𝗒 𝗍𝗈 𝗂𝖽𝖾𝗇𝗍𝗂𝖿𝗒 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗐𝗁𝗂𝖼𝗁 𝗂𝗌 𝗍𝗈 𝗋𝖾𝖺𝗅𝗂𝗓𝖾 𝗍𝗁𝖺𝗍 𝗐𝗁𝖾𝗇 𝗐𝖾 𝗁𝖺𝗏𝖾 𝖺 𝗠𝗮𝘁𝗰𝗵 𝗐𝖾 𝖽𝗈𝗇'𝗍 𝗂𝗇𝖼𝗋𝖾𝖺𝗌𝖾 𝗍𝗁𝖾 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗆𝗈𝗏𝗂𝗇𝗀 𝖽𝗂𝖺𝗀𝗈𝗇𝖺𝗅𝗅𝗒. 𝖶𝗁𝖾𝗋𝖾𝖺𝗌 𝗐𝗁𝖾𝗇 𝗂𝗍'𝗌 𝖺 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵 𝗐𝖾 𝗇𝖾𝖾𝖽 𝖺 𝗥𝗲𝗽𝗹𝗮𝗰𝗲𝗺𝗲𝗻𝘁 𝗈𝗉𝖾𝗋𝖺𝗍𝗂𝗈𝗇 𝗍𝗈 𝗆𝗈𝗏𝖾 𝖽𝗈𝗐𝗇 𝖽𝗂𝖺𝗀𝗈𝗇𝖺𝗅𝗅𝗒. 𝖨𝗇 𝗍𝗁𝗂𝗌 𝗉𝖺𝗋𝗍𝗂𝖼𝗎𝗅𝖺𝗋 𝖼𝖺𝗌𝖾 𝗒𝗈𝗎 𝖼𝖺𝗇 𝗍𝗋𝖺𝗏𝖾𝗋𝗌𝖾 𝖿𝗋𝗈𝗆

𝗗[ 𝟢 ][ 𝟢 ] 𝗍𝗈

𝗗[ 𝟤 ][ 𝟤 ]

𝖻𝗒:

𝖺. 𝖳𝗐𝗈 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵𝖾𝗌 𝗍𝗁𝖺𝗍 𝖻𝗈𝗍𝗁 𝗆𝗈𝗏𝖾 𝖽𝗂𝖺𝗀𝗈𝗇𝖺𝗅𝗅𝗒 𝖽𝗈𝗐𝗇 𝖻𝗒 𝟣 𝗉𝗈𝗂𝗇𝗍 𝖾𝖺𝖼𝗁 (𝖺𝖽𝖽𝗂𝗇𝗀 𝗍𝗈 𝖺 𝗍𝗈𝗍𝖺𝗅 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗈𝖿 𝗍𝗐𝗈). 𝖫𝗈𝗈𝗄 𝖺𝗍 𝗍𝗁𝖾 𝗉𝖺𝗍𝗁 𝗌𝗁𝗈𝗐𝗇 𝗎𝗌𝗂𝗇𝗀 𝗍𝗁𝖾 𝖺𝗅𝗂𝗀𝗇𝗆𝖾𝗇𝗍 𝗈𝗇 𝗍𝗁𝖾 𝗍𝗈𝗉 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗂𝗀𝗎𝗋𝖾. 𝖸𝗈𝗎 𝖼𝖺𝗇 𝗌𝖾𝖾 𝗍𝗁𝖺𝗍 𝗂𝗇 𝗍𝗁𝗂𝗌 𝗉𝖺𝗍𝗁 𝗍𝗁𝖾𝗋𝖾'𝗌 𝗇𝗈 𝖨𝗇𝗌𝖾𝗋𝗍𝗂𝗈𝗇𝗌 𝗈𝗋 𝗗𝗲𝗹𝗲𝘁𝗶𝗼𝗻𝗌 𝗐𝖾 𝗁𝖺𝗏𝖾 𝗗𝗜 𝗈𝗇 𝗍𝗁𝖾 𝗍𝗈𝗉 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾 𝖺𝗇𝖽 𝗘𝗗 𝗈𝗇 𝗍𝗁𝖾 𝖻𝗈𝗍𝗍𝗈𝗆 𝗈𝗇𝖾. 𝖠𝗇𝖽 𝗍𝗁𝖾𝗒 𝖺𝗋𝖾 𝖻𝗈𝗍𝗁 𝗠𝗶𝘀𝗺𝗮𝘁𝗰𝗵𝖾𝗌.

𝖻. 𝖳𝗁𝗂𝗌 𝗂𝗌 𝖺 𝗍𝗁𝗋𝖾𝖾 𝗌𝗍𝖾𝗉 𝗉𝖺𝗍𝗁 𝗂𝗇𝗏𝗈𝗅𝗏𝗂𝗇𝗀 𝖺𝗇 𝖨𝗇𝗌𝖾𝗋𝗍𝗂𝗈𝗇, 𝖬𝖺𝗍𝖼𝗁 𝖺𝗇𝖽 𝗗𝗲𝗹𝗲𝘁𝗲 𝗌𝗍𝗂𝗅𝗅 𝖺𝖽𝖽𝗂𝗇𝗀 𝗍𝗈 𝖺 𝗍𝗈𝗍𝖺𝗅 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗈𝖿 𝟤. 𝖸𝗈𝗎 𝖿𝗂𝗋𝗌𝗍 𝖨𝗇𝗌𝖾𝗋𝗍 𝗘 𝖺𝗍 𝗍𝗁𝖾 𝖻𝖾𝗀𝗂𝗇𝗇𝗂𝗇𝗀 𝗈𝖿 𝗍𝗁𝖾 𝖾𝗆𝗉𝗍𝗒 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝗉𝗋𝖾𝖿𝗂𝗑. 𝖳𝗁𝗂𝗌 𝗅𝖾𝖺𝖽𝗌 𝗍𝗈 𝖺𝗇 𝖺𝗎𝗍𝗈𝗆𝖺𝗍𝗂𝖼 𝗠𝗮𝘁𝗰𝗵 𝗈𝖿 𝗍𝗁𝖾 𝗍𝗐𝗈 𝗗 𝖼𝗁𝖺𝗋𝗌 𝗂𝗇 𝗍𝗁𝖾 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾𝗌. 𝖠𝗇𝖽 𝗍𝗁𝖾 𝖿𝗂𝗇𝖺𝗅 𝗈𝗉𝖾𝗋𝖺𝗍𝗂𝗈𝗇 𝗂𝗌 𝖺 𝗗𝗲𝗹𝗲𝘁𝗶𝗼𝗻 𝗈𝖿 𝗍𝗁𝖾 𝖨 𝖼𝗁𝖺𝗋 𝖿𝗋𝗈𝗆 𝗍𝗁𝖾 𝗗𝗜𝗦𝗧𝗔𝗡𝗖𝗘 𝗌𝖾𝗊𝗎𝖾𝗇𝖼𝖾.

𝖳𝖺𝗄𝖾 𝖺 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗈𝗇𝖾 𝗆𝗈𝗋𝖾 𝖻𝗂𝗀𝗀𝖾𝗋 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝖻𝖾𝗅𝗈𝗐. 𝖬𝖺𝗄𝖾 𝗌𝗎𝗋𝖾 𝗒𝗈𝗎 𝗐𝖺𝗅𝗄 𝗍𝗁𝗋𝗈𝗎𝗀𝗁 𝗍𝗁𝖾 𝗉𝖺𝗍𝗁𝗌 𝗆𝖾𝗍𝗂𝖼𝗎𝗅𝗈𝗎𝗌𝗅𝗒 𝖺𝗇𝖽 𝖼𝗈𝗇𝗏𝗂𝗇𝖼𝖾 𝗒𝗈𝗎𝗋𝗌𝖾𝗅𝖿 𝗍𝗁𝖺𝗍 𝖾𝖺𝖼𝗁 𝗈𝖿 𝗍𝗁𝖾 𝗉𝖺𝗍𝗁 𝗀𝗈𝗂𝗇𝗀 𝖿𝗋𝗈𝗆

𝗗[ 𝟢 ][ 𝟢 ] 𝗍𝗈

𝗗[ 𝟤 ][ 𝟪 ]

𝖽𝗈𝖾𝗌 𝗂𝗇𝖽𝖾𝖾𝖽 𝗅𝖾𝖺𝖽 𝗍𝗈 𝗍𝗁𝖾 𝗌𝖺𝗆𝖾 𝖿𝗂𝗇𝖺𝗅 𝖽𝗂𝗌𝗍𝖺𝗇𝖼𝖾 𝗌𝖼𝗈𝗋𝖾 𝗈𝖿 𝟪.

Edit Distance | Tracing Parent Pointers | One of the Sequence of Operations required to go from D[0][0] to D[2][8] shown on the top of the Figure
Edit Distance | Tracing Parent Pointers | Two other Sequence of Operations that take you from D[0][0] to D[2][8] shown on the left and bottom of the Figure
Edit Distance | Tracing Parent Pointers | Sequences of Operations that take you from D[0][0] to D[2][8]

𝖭𝗈𝗐 𝗅𝖾𝗍'𝗌 𝗍𝖺𝗄𝖾 𝖺 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗍𝗁𝖾 𝖼𝗈𝗆𝗉𝗅𝖾𝗍𝖾 𝗺𝗲𝗺𝗼 𝗍𝖺𝖻𝗅𝖾 𝖿𝗈𝗋 𝗍𝗁𝖾 𝗘𝗱𝗶𝘁 𝗗𝗶𝘀𝘁𝗮𝗻𝗰𝗲 𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝖻𝖾𝗅𝗈𝗐

Edit Distance | Tracing Parent Pointers | Complete Memo Table

𝖠𝗇𝖽 𝗅𝖾𝗍'𝗌 𝗍𝖺𝗄𝖾 𝖺 𝗅𝗈𝗈𝗄 𝖺𝗍 𝗈𝗇𝖾 𝖾𝗑𝖺𝗆𝗉𝗅𝖾 𝗉𝖺𝗍𝗁 𝖿𝗋𝗈𝗆

𝗗[ 𝟢 ][ 𝟢 ] 𝗍𝗈

𝗗[ 𝟩 ][ 𝟪 ]

𝗍𝗁𝖾 𝖿𝗂𝗇𝖺𝗅 𝗢𝗽𝘁𝗶𝗺𝗮𝗹 𝗔𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁

Edit Distance | Tracing Parent Pointers | Complete Example Path (in Dotted Black Arrows) and its Corresponding Sequence of Operations at the Bottom

𝖨𝗇 𝖼𝗈𝖽𝖾 / 𝗉𝗋𝖺𝖼𝗍𝗂𝖼𝖾 𝖿𝗈𝗋 𝗍𝗋𝖺𝖼𝗂𝗇𝗀 𝗉𝖺𝗋𝖾𝗇𝗍 𝗉𝗈𝗂𝗇𝗍𝖾𝗋𝗌 𝗒𝗈𝗎 𝗌𝗍𝖺𝗋𝗍 𝗍𝗁𝖾 𝖿𝗂𝗇𝖺𝗅 𝖾𝗇𝖽 𝗉𝗈𝗂𝗇𝗍 𝗈𝖿

𝗗[ 𝟩 ][ 𝟪 ] = 𝟧

𝖺𝗇𝖽 𝗐𝗈𝗋𝗄 𝗒𝗈𝗎𝗋 𝗐𝖺𝗒 𝖻𝖺𝖼𝗄𝗐𝖺𝗋𝖽𝗌 𝖺𝗇𝖽 𝗁𝖾𝗇𝖼𝖾 𝗍𝗁𝖾 𝗍𝖾𝗋𝗆 𝗍𝗋𝖺𝖼𝗂𝗇𝗀 (𝖻𝖺𝖼𝗄𝗍𝗋𝖺𝖼𝗄𝗂𝗇𝗀) 𝗉𝖺𝗋𝖾𝗇𝗍 𝗉𝗈𝗂𝗇𝗍𝖾𝗋𝗌. 𝖸𝗈𝗎 𝖼𝗈𝗎𝗅𝖽 𝗎𝗌𝖾 𝖾𝗂𝗍𝗁𝖾𝗋 𝖺 𝘀𝘁𝗮𝗰𝗸 𝗈𝗋 𝖺 𝗾𝘂𝗲𝘂𝗲 𝗍𝗈 𝗁𝗈𝗅𝖽 𝗍𝗁𝖾 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝗈𝖿 𝗼𝗽𝗲𝗿𝗮𝘁𝗶𝗼𝗻𝘀. 𝖴𝗌𝗂𝗇𝗀 𝖺 𝗾𝘂𝗲𝘂𝗲 𝗐𝗈𝗎𝗅𝖽 𝖼𝗈𝗋𝗋𝖾𝗌𝗉𝗈𝗇𝖽 𝗍𝗈 𝘄𝗮𝗹𝗸𝗶𝗻𝗴 𝗯𝗮𝗰𝗸𝘄𝗮𝗿𝗱𝘀 𝖿𝗋𝗈𝗆 𝗍𝗁𝖾 𝗳𝗶𝗻𝗮𝗹 𝗮𝗹𝗶𝗴𝗻𝗺𝗲𝗻𝘁 𝖺𝗅𝗅 𝗍𝗁𝖾 𝗐𝖺𝗒 𝖻𝖺𝖼𝗄 𝗍𝗈 𝗍𝗁𝖾 𝗲𝗺𝗽𝘁𝘆 𝗽𝗿𝗲𝗳𝗶𝘅𝖾𝗌 𝖺𝗍

𝗗[ 𝟢 ][ 𝟢 ]

𝖶𝗁𝖾𝗋𝖾𝖺𝗌 𝖺 𝘀𝘁𝗮𝗰𝗸 𝗐𝗈𝗎𝗅𝖽 𝖼𝗈𝗋𝗋𝖾𝗌𝗉𝗈𝗇𝖽 𝗍𝗈 𝘄𝗮𝗹𝗸𝗶𝗻𝗴 𝗳𝗼𝗿𝘄𝗮𝗿𝗱 𝖿𝗋𝗈𝗆 𝗍𝗁𝖾 𝗲𝗺𝗽𝘁𝘆 𝗽𝗿𝗲𝗳𝗶𝘅𝖾𝗌 𝖺𝗅𝗅 𝗍𝗁𝖾 𝗐𝖺𝗒 𝗍𝗈 𝗍𝗁𝖾 𝖼𝗈𝗆𝗉𝗅𝖾𝗍𝖾 𝗌𝗍𝗋𝗂𝗇𝗀𝗌. 𝖠𝗇𝖽 𝗂𝗇 𝗍𝗁𝖾 𝖼𝖺𝗌𝖾 𝗈𝖿 𝗦𝗵𝗼𝗿𝘁𝗲𝘀𝘁 𝗣𝗮𝘁𝗵 𝗂𝗇 𝗗𝗔𝗚 𝗒𝗈𝗎 𝗌𝗍𝖺𝗋𝗍 𝖺𝗍 𝗍𝗁𝖾 𝗌𝗍𝖺𝗋𝗍𝗂𝗇𝗀 𝗉𝗈𝗂𝗇𝗍 𝖺𝗇𝖽 𝗐𝗈𝗋𝗄 𝗒𝗈𝗎𝗋 𝗐𝖺𝗒 𝖿𝗈𝗋𝗐𝖺𝗋𝖽 𝖻𝗒 𝖺𝗌𝗌𝗂𝗀𝗇𝗂𝗇𝗀 𝖤𝖽𝗀𝖾 𝖶𝖾𝗂𝗀𝗁𝗍𝗌.

𝖳𝗁𝗂𝗌 𝖺𝗋𝗍𝗂𝖼𝗅𝖾 𝗁𝖺𝗌 𝖺𝗅𝗋𝖾𝖺𝖽𝗒 𝖻𝖾𝖼𝗈𝗆𝖾 𝗊𝗎𝗂𝗍𝖾 𝖺 𝗅𝗈𝗇𝗀 𝗋𝖾𝖺𝖽 𝗌𝗈 𝖨'𝗅𝗅 𝗌𝗍𝗈𝗉 𝗁𝖾𝗋𝖾 𝖿𝗈𝗋 𝗇𝗈𝗐 𝖺𝗇𝖽 𝗐𝖾'𝗅𝗅 𝖼𝗈𝗏𝖾𝗋 𝗍𝗁𝖾 𝖿𝗈𝗅𝗅𝗈𝗐𝗂𝗇𝗀 𝖺𝗌 𝗉𝗋𝗈𝗆𝗂𝗌𝖾𝖽 𝗂𝗇 𝖺 𝗌𝗎𝖻𝗌𝖾𝗊𝗎𝖾𝗇𝗍 𝖺𝗋𝗍𝗂𝖼𝗅𝖾 𝗌𝗈𝗈𝗇. 𝖲𝗍𝖺𝗒 𝗍𝗎𝗇𝖾𝖽.

𝟤. 𝗦𝘂𝗯𝘀𝘁𝗿𝗶𝗻𝗴𝗌

𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗣𝗮𝗹𝗶𝗻𝗱𝗿𝗼𝗺𝗶𝗰 𝗦𝘂𝗯𝘀𝘁𝗿𝗶𝗻𝗴

𝗂𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗣𝗮𝗹𝗶𝗻𝗱𝗿𝗼𝗺𝗶𝗰 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲

𝗂𝗂𝗂. 𝗠𝗮𝘅 𝗦𝘂𝗺 𝗦𝘂𝗯𝗮𝗿𝗿𝗮𝘆

𝗂𝗏. 𝗠𝗮𝘅 𝗣𝗿𝗼𝗱𝘂𝗰𝘁 𝗦𝘂𝗯𝗮𝗿𝗿𝗮𝘆

𝟥. 𝗔𝗱𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗖𝗼𝗻𝘀𝘁𝗿𝗮𝗶𝗻𝘁𝘀 (𝖲𝗎𝖻𝗉𝗋𝗈𝖻𝗅𝖾𝗆 𝗦𝗽𝗮𝗰𝗲 𝖺𝗇𝖽 𝗥𝗲𝗹𝗮𝘁𝗶𝗼𝗻𝘀𝗵𝗶𝗽 𝗘𝘅𝗽𝗮𝗻𝘀𝗶𝗼𝗻)

𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 (𝖦𝗈𝗈𝖽) 𝗣𝗮𝗹𝗶𝗻𝗱𝗿𝗼𝗺𝗶𝗰 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 𝖨𝖨

𝗂𝗂. 𝗟𝗼𝗻𝗴𝗲𝘀𝘁 𝗜𝗻𝗰𝗿𝗲𝗮𝘀𝗶𝗻𝗴 𝗦𝘂𝗯𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲


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