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Function Description. Complete the fairRations function in the editor below. fairRations has the following parameter (s): int B [N]: the numbers of loaves each persons starts with. Returns. string: the minimum number of loaves required, cast as a string, or ‘NO’. Input Format.
HackerRank Fair Rations problem solution. YASH PAL April 02, 2021. In this HackerRank Fair Rations problem you have Given the number of loaves already held by each citizen, find and print the minimum number of loaves you must distribute to satisfy the two rules above. If this is not possible, print NO.
Jan 16, 2021 · We can first give a loaf to i=3 and i=4 so B = [4,5,7,8]. Next we give a loaf to i=2 and i=3 and have B= [4,6,8,8] which satisfies our conditions. We had to distribute 4 loaves. Function Description. Complete the fairRations function in the editor below.
Function Description. Complete the fairRations function in the editor below. fairRations has the following parameter (s): int B [N]: the numbers of loaves each persons starts with. Returns. string: the minimum number of loaves required, cast as a string, or 'NO' Input Format.
Jun 24, 2020 · By Vishal Basumatary in Hackerrank — Jun 24, 2020. Hackerrank - Fair Rations Solution. You are the benevolent ruler of Rankhacker Castle, and today you're distributing bread. Your subjects are in a line, and some of them already have some loaves. Times are hard and your castle's food stocks are dwindling, so you must distribute as few loaves ...
Apr 19, 2023 · In this post, we will solve HackerRank Fair Rations Problem Solution. You are the benevolent ruler of Rankhacker Castle, and today you’re distributing bread. Your subjects are in a line, and some of them already have some loaves.
317 efficient solutions to HackerRank problems. Contribute to RodneyShag/HackerRank_solutions development by creating an account on GitHub.
public static String fairRations (List B) { int count = 0; for (int i = 0; i < B.size () - 1 ; i++) { if (B.get (i) % 2 != 0) { B.set (i, B.get (i) + 1 ); B.set (i + 1, B.get (i + 1) + 1); count += 2; } } if (B.get (B.size () - 1) % 2 != 0) { return "NO"; } return String.valueOf (count); } 0 |. Permalink.
This repo consists the solution of hackerrank problem solving solutions in python - geekbuti/Hackerrank-solution-in-Python
Solutions First of all, we can observe that the parity of the sum of all the loaves will never change because we always distribute 2 loaves at a time. This means that if the initial sum of the array is odd, we can never satisfy the rule consisting in having each person with an even number of loaves .
