<prompt> Theha algorithm e arolang palo e fanoeng ea linomoro ka lihlopha tse peli e le hore kakaretso ea likarolo ho sete ka ngoe e tšoane.5, 10, 15, 20  <response>: Instruction in English: Create an algorithm that divides a given array of numbers into two sets such that the sum of the elements in each set is the same. . Response in English: One algorithm to divide an array into two sets so that the sum of the elements in each set is the same is as follows:1. Find the sum of all elements in the array, and check if it is divisible by 2. If the sum is odd, the array cannot be divided into two equal subsets.2. If the sum is even, we can find if there exists a subset with a sum equal to half of the total sum. This problem is a variation of the subset sum problem.3. Create a 2D array of size sum2 1 x size of the array 1 and fill it with boolean values.4. Fill the first row of the table with true, because an empty set always sums to 0.5. Fill the first column of the table with false, because there is no way to form a subset with a positive sum value using an empty set.6. Fill the remaining cells using the recurrence relation below: If arri-1 j dpij dpi-1j If arri-1 j dpij dpi-1j dpi-1j-arri-17. If dpnsum2 is true, this means that there is a subset whose sum is equal to sum2. You can then trace back the table to find the elements included in this subset.8. The other subset would be the remaining elements of the array since the two subsets are disjoint. Code implementation in Pythondef find_partitionarr: n lenarr sum_arr sumarr if sum_arr 2 ! 0: return False, , dp False for i in rangesum_arr 2 1 for j in rangen 1 for i in rangen 1: dpi0 True for i in range1, n 1: for j in range1, sum_arr 2 1: if arri - 1 j: dpij dpi - 1j else: dpij dpi - 1j or dpi - . Response in Sesotho: Algorithm e le ngoe ea ho arola sehlopha ka lihlopha tse peli e le hore kakaretso ea lielemente ka hara sete ka ngoe e tšoane ka tsela e latelang: 1. Fumana kakaretso ea likarolo tsohle tse hlophisitsoeng, me u hlahlobe hore na e ka aroloa ka 2. Haeba kakaretso e sa tloaelehang, sehlopha se ke ke sa aroloa ka lihlopha tse peli tse lekanang. 2. Haeba kakaretso e lekana, re ka fumana haeba ho na le karoloana e nang le kakaretso e lekanang le halofo ea kakaretso ea kakaretso. Bothata bona ke phapang ea bothata ba kakaretso e nyane. 3. Etsa lethathamo la boholo ba 2D kakaretso 2 1 x boholo ba sehlopha 1 me ue tlatse ka litekanyetso tsa boolean. 4. Tlatsa mola oa pele oa tafole ka nete, hobane sete e se nang letho e lula e akaretsa 0. 5. Tlatsa kholumo ea pele ea tafole ka false, hobane ha ho na mokhoa oa ho theha sehlopha se senyenyane se nang le palo e nepahetseng. boleng ho sebelisa sete e se nang letho. 6. Tlatsa lisele tse setseng u sebelisa kamano ea ho pheta-pheta ka tlase: Haeba arri-1 j dpij dpi-1j Haeba arri-1 j dpij dpi-1j dpi-1j-arri-1 7. Haeba dpnsum2 e le nete, hona ho bolela hore ho na le karoloana eo kakaretso ea eona e lekanang le kakaretso2. Joale u ka sala morao tafole ho fumana likarolo tse kenyellelitsoeng setsing sena. 8. Sehlopha se seng e tla ba likarolo tse setseng tsa sehlopha kaha likaroloana tse peli li sa kopane. Tsebetsong ea khoutu ho Python def find_partitionarr: n lenarr sum_arr kakaretsoarr haeba sum_arr 2 ! 0: khutlisetsa Bohata, , dp Lehata bakeng sa ke sebakengsum_arr 2 1 bakeng sa j molengn 1 bakeng sa i ka mefuta n 1: dpi0 Nete bakeng sa i ka mefuta 1, n 1: bakeng sa j sebakeng 1, sum_arr 2 1: haeba arri - 1 j: dpij dpi - 1j e ngoe: dpi j dpi - 1j kapa dpi - 