In this article, we will deal with the Bridge Building<\/strong> problem, which is one of the coding test problems using JavaScript. This problem utilizes the concept of combinations and has many applications. We will examine the problem definition, various approaches, and the solution process in great detail.<\/p>\n Problem:<\/strong> There are two islands in a river. We want to build a bridge between the two islands. Given that we can build N bridges, we need to determine how many different ways we can construct these bridges. However, the bridges must not interfere with each other, and there are designated positions for the bridges on each island.<\/p>\n This problem serves as a foundation for understanding the properties of combinations. The positions of the bridges must be placed at fixed positions on the two islands, so the selection of bridges must not interfere with each other. We will refer to the two islands where the bridges will be placed as There are various approaches to solving combination problems. In the case of this problem, it is advisable to use the Dynamic Programming method. First, we will set basic situations as needed and develop more complex situations based on those.<\/p>\n To solve the bridge building problem using dynamic programming, we can set up the following recurrence relation:<\/p>\n Here, Now, based on the above recurrence relation, let\u2019s implement the algorithm in JavaScript. Here is an example of the function:<\/p>\n The above function works through the following steps:<\/p>\n The time complexity of this algorithm is O(N). This is because it calculates the result by traversing the array once. The space complexity is also O(N) since an array is used to store the number of ways depending on the number of bridges.<\/p>\n Through this tutorial, we have learned how to solve the bridge building problem using JavaScript. This problem greatly aids in understanding the concept of combinations and optimization approaches using dynamic programming. We hope to expand basic programming skills and develop the ability to utilize various algorithms through such problems. We look forward to solving more algorithm problems together in the future!<\/p>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"Problem Definition<\/h2>\n
Example Input\/Output<\/h3>\n
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Problem Analysis<\/h2>\n
A<\/code> and B<\/code>. The positions for placing bridges on each island can be represented by indices starting from 0.<\/p>\nApproach<\/h2>\n
Dynamic Programming Approach<\/h3>\n
count(N) = count(N-1) + count(N-2)<\/code><\/pre>\ncount(N)<\/code> represents the number of ways to place N bridges. The meaning of this equation is as follows:<\/p>\n\n
N-1<\/code> bridges, adding the last bridge (only one bridge is added).<\/li>\nN-2<\/code> bridges, adding two new bridges.<\/li>\n<\/ul>\nAlgorithm Implementation<\/h2>\n
\nfunction countWays(n) {\n if (n === 0) return 1; \/\/ Base case: not placing any bridges\n if (n === 1) return 1; \/\/ Base case: placing one bridge\n\n let dp = new Array(n + 1);\n dp[0] = 1;\n dp[1] = 1;\n\n for (let i = 2; i <= n; i++) {\n dp[i] = dp[i - 1] + dp[i - 2];\n }\n\n return dp[n];\n}\n \nlet n = 3;\nconsole.log(countWays(n)); \/\/ Result: 5\n <\/code><\/pre>\nCode Explanation<\/h3>\n
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n<\/code> is 0, it returns 1 to handle the base case.<\/li>\nn<\/code> is 1, it also returns 1 to handle the case of placing only one bridge.<\/li>\ndp<\/code> based on the number of bridges, and uses the recurrence relation to calculate the number of ways to place the bridges.<\/li>\n<\/ul>\nTime Complexity<\/h2>\n
Conclusion<\/h2>\n