Tasks Details
medium
1.
Multivitamin
There are N glasses of different capacities, each of them containing a different amount of a unique kind of juice. Calculate the maximum number of kinds of juice that can be mixed in a single glass.
Task Score
100%
Correctness
100%
Performance
100%
Rick is really fond of fruit juices, but he is bored of their traditional flavours. Therefore, he has decided to mix as many of them as possible to obtain something entirely new as a result.
He has N glasses, numbered from 0 to N-1, each containing a different kind of juice. The J-th glass has capacity[J] units of capacity and contains juice[J] units of juice. In each glass there is at least one unit of juice.
Rick want to create a multivitamin mix in one of the glasses. He is going to do it by pouring juice from several other glasses into the chosen one. Each of the used glasses must be empty at the end (all of the juice from each glass has to be poured out).
What is the maximum number of flavours that Rick can mix?
Write a function:
function solution(juice, capacity);
that, given arrays juice and capacity, both of size N, returns the maximum number of flavours that Rick can mix in a single glass.
Examples:
1. Given juice = [10, 2, 1, 1] and capacity = [10, 3, 2, 2], your function should return 2. Rick can pour juice from the 3rd glass into the 2nd one.
2. Given juice = [1, 2, 3, 4] and capacity = [3, 6, 4, 4], your function should return 3. Rick can pour juice from the 0th and 2nd glasses into the 1st one.
3. Given juice = [2, 3] and capacity = [3, 4], your function should return 1. No matter which glass he chooses, Rick cannot pour juice from the other one into it. The maximum number of flavours in the chosen glass is 1.
4. Given juice = [1, 1, 5] and capacity = [6, 5, 8], your function should return 3. Rick can mix all juices in the 2nd glass.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [2..100,000];
- each element of arrays juice and capacity is an integer within the range [1..1,000,000,000];
- arrays juice and capacity have the same length, equal to N;
- for each J juice[J] ≤ capacity[J].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used JavaScript
Time spent on task 6 minutes
Notes
not defined yet
Code: 08:36:59 UTC,
java,
autosave
Code: 08:37:10 UTC,
js,
autosave
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = -1;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
lastTestesGlass = i;
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
maxJuiceFlavors++;
if(i + 1 == glasses.length) lastTestesGlass = glasses.length;// Edge case where all glasses were added
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) {
return maxJuiceFlavors + 1;
}
}
return maxJuiceFlavors;
}
Code: 08:38:54 UTC,
js,
autosave
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = -1;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
maxJuiceFlavors++;
lastTestesGlass = i + 1;
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) {
return maxJuiceFlavors + 1;
}
}
return maxJuiceFlavors;
}
Code: 08:38:58 UTC,
js,
verify,
result: Failed
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = -1;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
maxJuiceFlavors++;
lastTestesGlass = i + 1;
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) {
return maxJuiceFlavors + 1;
}
}
return maxJuiceFlavors;
}
Analysis
expand all
Example tests
1.
0.072 s
OK
1.
0.072 s
OK
1.
0.072 s
RUNTIME ERROR,
tested program terminated with exit code 1
stderr:
TypeError: Cannot read property '1' of undefined
at solution (solution.js:45:28)
at solutionWrapper (/tmp/exec.js:412:28)
at Promise.resolve.then (/tmp/exec.js:438:24)
at <anonymous>
at process._tickCallback (internal/process/next_tick.js:188:7)
at Function.Module.runMain (module.js:686:11)
at startup (bootstrap_node.js:187:16)
at bootstrap_node.js:608:3
1.
0.072 s
OK
Code: 08:42:29 UTC,
js,
autosave
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = 0;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
lastTestesGlass = i + 1;
maxJuiceFlavors++;
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) return maxJuiceFlavors + 1;
}
return maxJuiceFlavors;
}
Code: 08:42:30 UTC,
js,
verify,
result: Passed
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = 0;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
lastTestesGlass = i + 1;
maxJuiceFlavors++;
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) return maxJuiceFlavors + 1;
}
return maxJuiceFlavors;
}
Analysis
Code: 08:42:39 UTC,
js,
verify,
result: Passed
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = 0;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
lastTestesGlass = i + 1;
maxJuiceFlavors++;
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) return maxJuiceFlavors + 1;
}
return maxJuiceFlavors;
}
Analysis
Code: 08:42:41 UTC,
js,
final,
score:
100
function solution(juice, capacity) {
let maxJuiceFlavors = 1;// number of diferent flavors
let currJuiceInGlass = 0;// the sum of juice quantities
let targetGlass = -1;
let lastTestesGlass = 0;
let currCapacity = 0;
let glasses = [];
// Find the glass with the max empty space
for (let i = 0; i < juice.length; i++) {
let j = juice[i];
let c = capacity[i];
let e = c - j;
glasses.push([j, c]);
if(e > currCapacity){
currCapacity = e;
targetGlass = i;
}
}
if(currCapacity < 1) return maxJuiceFlavors;
currJuiceInGlass = glasses[targetGlass][0];
// Remove the targetGlass from the new glasses array
glasses.splice(targetGlass,1);
// Sort the remaining glasses
glasses.sort((g1, g2) => g1[0] == g2[0] ? g1[1] - g2[1] : g1[0] - g2[0]);
// Pour every other juice using the target glass capacity
for(let i = 0; i < glasses.length; i++){
currCapacity -= glasses[i][0];
if(currCapacity < 0) break;
currJuiceInGlass += glasses[i][0];
lastTestesGlass = i + 1;
maxJuiceFlavors++;
}
// Find edge case where one other glass can fit more juice
for(let i = 0; i < glasses.length - lastTestesGlass; i++){
let currGlass = i + lastTestesGlass;
if(glasses[currGlass][1] - glasses[currGlass][0] >= currJuiceInGlass) return maxJuiceFlavors + 1;
}
return maxJuiceFlavors;
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N * log(N))
expand all
Correctness tests
1.
0.076 s
OK
2.
0.076 s
OK
3.
0.076 s
OK
4.
0.076 s
OK
5.
0.076 s
OK
6.
0.076 s
OK
7.
0.076 s
OK
1.
0.076 s
OK
2.
0.072 s
OK
3.
0.076 s
OK
4.
0.076 s
OK
1.
0.072 s
OK
2.
0.076 s
OK
3.
0.076 s
OK
4.
0.072 s
OK
5.
0.072 s
OK
6.
0.072 s
OK
not_biggest_capacity
Tests in which the glass with the biggest capacity is not the one that should be picked as the base.
Tests in which the glass with the biggest capacity is not the one that should be picked as the base.
✔
OK
1.
0.072 s
OK
2.
0.076 s
OK
not_biggest_capacity_left
Tests in which the glass with the largest empty space is not the picked as the base.
Tests in which the glass with the largest empty space is not the picked as the base.
✔
OK
1.
0.076 s
OK
2.
0.076 s
OK
3.
0.072 s
OK
4.
0.076 s
OK
1.
0.076 s
OK
2.
0.076 s
OK
3.
0.076 s
OK
1.
0.072 s
OK
2.
0.076 s
OK
3.
0.072 s
OK
4.
0.076 s
OK
1.
0.072 s
OK
2.
0.076 s
OK
expand all
Performance tests
1.
0.092 s
OK
2.
0.128 s
OK
3.
0.124 s
OK
1.
0.096 s
OK
2.
0.400 s
OK
3.
0.392 s
OK
4.
0.376 s
OK
5.
0.308 s
OK
1.
0.076 s
OK
2.
0.072 s
OK
3.
0.072 s
OK
4.
0.196 s
OK
5.
0.220 s
OK
1.
0.412 s
OK
2.
0.428 s
OK
1.
0.404 s
OK
2.
0.208 s
OK
3.
0.384 s
OK
4.
0.364 s
OK
5.
0.396 s
OK