Test results - Codility

Tasks Details

medium

1. NumberOfDiscIntersections

Compute the number of intersections in a sequence of discs.

Task Score

100%

Correctness

100%

Performance

100%

We draw N discs on a plane. The discs are numbered from 0 to N − 1. An array A of N non-negative integers, specifying the radiuses of the discs, is given. The J-th disc is drawn with its center at (J, 0) and radius A[J].

We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).

The figure below shows discs drawn for N = 6 and A as follows:

A[0] = 1 A[1] = 5 A[2] = 2 A[3] = 1 A[4] = 4 A[5] = 0

There are eleven (unordered) pairs of discs that intersect, namely:

discs 1 and 4 intersect, and both intersect with all the other discs;

disc 2 also intersects with discs 0 and 3.

Write a function:

function solution(A);

that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.

Given array A shown above, the function should return 11, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [0..2,147,483,647].

Solution

Programming language used JavaScript

Time spent on task 3 minutes

Notes

not defined yet

Task timeline

Code: 08:14:39 UTC, java, autosave

show code in pop-up

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// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // Implement your solution here
    }
}

Code: 08:14:40 UTC, java, autosave

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // Implement your solution here
    }
}

Code: 08:14:50 UTC, js, autosave

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function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        if(total - no_intersect > 10000000) return -1
    }
    return total - no_intersect
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

Code: 08:15:12 UTC, js, autosave

show code in pop-up

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function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        if((total - no_intersect) > 10000000) return -1
    }
    return total - no_intersect
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

Code: 08:15:25 UTC, js, verify, result: Passed

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function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        if((total - no_intersect) > 10000000) return -1
    }
    return total - no_intersect
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

User test case 1:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Analysis

expand all Example tests

▶

example1
example test

✔

0.056 s

expand all User tests

▶

test_data1

✔

0.056 s

OK
function result: 0

Code: 08:15:44 UTC, js, autosave

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function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        
    }
    return total - no_intersect
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

Code: 08:15:55 UTC, js, autosave

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        
    }
    return (total - no_intersect > 10000000) return -1
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

Code: 08:16:14 UTC, js, autosave

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        
    }
    var 
    return (total - no_intersect > 10000000) ? -1:total - no_intersect
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

Code: 08:16:36 UTC, js, verify, result: Passed

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        
    }
    var intersecting = total - no_intersect
    return intersecting > 10000000 ? -1: intersecting
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

User test case 1:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Analysis

expand all Example tests

▶

example1
example test

✔

0.060 s

expand all User tests

▶

test_data1

✔

0.052 s

OK
function result: 0

Code: 08:16:43 UTC, js, verify, result: Passed

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        
    }
    var intersecting = total - no_intersect
    return intersecting > 10000000 ? -1: intersecting
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

User test case 1:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Analysis

expand all Example tests

▶

example1
example test

✔

0.064 s

expand all User tests

▶

test_data1

✔

0.056 s

OK
function result: 0

Code: 08:16:48 UTC, js, final, score: 100

show code in pop-up

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function solution(A) {
    var range = []
    var total = A.length*(A.length-1)/2
    var no_intersect = 0
    for(var i=0;i<A.length;i++){
        var count = binarySearch(range, i-A[i])
        no_intersect += count
        //console.log(range,count, i-A[i])
        range.splice(binarySearch(range, i+A[i]), 0, i+A[i])
        
    }
    var intersecting = total - no_intersect
    return intersecting > 10000000 ? -1: intersecting
}
//在一個已排序的陣列裡，取得newValue應插入的index
function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            //在這邊確認拿到的是"第一個"大於等於目標值的數字
            while(mid > 0 && arr[mid-1] === target){
                mid--
            }
            return mid;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return left
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N*log(N)) or O(N)

expand all Example tests

▶

example1
example test

✔

0.052 s

expand all Correctness tests

▶

simple1

✔

0.052 s

▶

simple2

✔

0.052 s

▶

simple3

✔

0.056 s

▶

extreme_small
empty and [10]

✔

0.052 s

▶

small1

✔

0.052 s

▶

small2

✔

0.052 s

▶

small3

✔

0.052 s

▶

overflow
arithmetic overflow tests

✔

0.056 s

0.052 s

expand all Performance tests

▶

medium1

✔

0.060 s

▶

medium2

✔

0.060 s

▶

medium3

✔

0.068 s

▶

medium4

✔

0.072 s

▶

10M_intersections
10.000.000 intersections

✔

0.076 s

▶

big1

✔

0.100 s

0.056 s

▶

big2

✔

0.132 s

0.052 s

▶

big3
[0]*100.000

✔

0.104 s

0.052 s