Don't say so much, just coding...
John and Mary want to travel between a few towns A, B, C ... Mary has on a sheet of paper a list of distances between these towns. ls = [50, 55, 57, 58, 60]
. John is tired of driving and he says to Mary that he doesn't want to drive more than t = 174
miles and he will visit only 3
towns.
Which distances, hence which towns, they will choose so that the sum of the distances is the biggest possible to please Mary and John?
Example:
With list ls
and 3 towns to visit they can make a choice between: [50,55,57],[50,55,58],[50,55,60],[50,57,58],[50,57,60],[50,58,60],[55,57,58],[55,57,60],[55,58,60],[57,58,60]
.
The sums of distances are then: 162, 163, 165, 165, 167, 168, 170, 172, 173, 175
.
The biggest possible sum taking a limit of 174 into account is then 173 and the distances of the 3 corresponding towns is [55, 58, 60]
.
The function chooseBestSum (or choose_best_sum
or ... depending on the language) will take as parameters t (maximum sum of distances, integer >= 0), k
(number of towns to visit, k >= 1) and ls
(list of distances, all distances are positive or null integers and this list has at least one element). The function returns the "best" sum ie the biggest possible sum of k distances less than or equal to the given limit t, if that sum exists, or otherwise nil, null, None, Nothing, depending on the language. With C++, C, Rust, Swift, Go, Kotlin return -1
.
Examples:
ts = [50, 55, 56, 57, 58] choose_best_sum(163, 3, ts) -> 163
xs = [50] choose_best_sum(163, 3, xs) -> nil (or null or ... or -1 (C++, C, Rust, Swift, Go)
ys = [91, 74, 73, 85, 73, 81, 87] choose_best_sum(230, 3, ys) -> 228
def choose_best_sum(t, k, ls)
# your code
end
Test.describe("choose_best_sum") do
Test.it("Basic Tests") do
ts = [50, 55, 56, 57, 58]
Test.assert_equals(choose_best_sum(163, 3, ts), 163)
ts = [50]
Test.assert_equals(choose_best_sum(163, 3, ts), nil)
ts = [91, 74, 73, 85, 73, 81, 87]
Test.assert_equals(choose_best_sum(230, 3, ts), 228)
end
end
function chooseBestSum(t, k, ls) {
// your code
}
Test.describe("chooseBestSum",function() {
Test.it("Basic tests ",function() {
var ts = [50, 55, 56, 57, 58]
Test.assertEquals(chooseBestSum(163, 3, ts), 163)
ts = [50]
Test.assertEquals(chooseBestSum(163, 3, ts), null)
ts = [91, 74, 73, 85, 73, 81, 87]
Test.assertEquals(chooseBestSum(230, 3, ts), 228)
})})
想法(1): 以 k
個 towns 來作為陣列,並且加總然後選擇小於 t
miles 的數字,並且找出最大值
想法(2): 這邊查到 combination
這個 ruby 的方法減少很多麻煩... 寫了好幾題覺得 ruby 真好 ·_·
圖片來源:Unsplash Ergita Sela
# Solution 1
def choose_best_sum(t, k, ls)
ls.combination(k)
.map{ |array| array.reduce(:+) }
.select{ |sum| sum <= t }
.max
end
# Solution 2
def choose_best_sum(t, k, ls)
ls.combination(k).map(&:sum).reject { |n| n > t }.max
end
// Solution 1
function chooseBestSum(t, k, ls) {
var biggestCount = 0;
var recurseTowns = function(townsSoFar, lastIndex) {
townsSoFar = townsSoFar || [];
if (townsSoFar.length === k) {
var sumDistance = townsSoFar.reduce((a,b)=>a+b);
if (sumDistance <= t && sumDistance > biggestCount) {
biggestCount = sumDistance;
}
return;
}
for (var i = lastIndex + 1 || 0; i < ls.length; i++) {
recurseTowns(townsSoFar.concat(ls[i]), i);
}
}
recurseTowns();
return biggestCount || null;
}