Tasks Details
easy
1.
FrogRiverOne
Find the earliest time when a frog can jump to the other side of a river.
Task Score
100%
Correctness
100%
Performance
100%
A small frog wants to get to the other side of a river. The frog is initially located on one bank of the river (position 0) and wants to get to the opposite bank (position X+1). Leaves fall from a tree onto the surface of the river.
You are given an array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in seconds.
The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X (that is, we want to find the earliest moment when all the positions from 1 to X are covered by leaves). You may assume that the speed of the current in the river is negligibly small, i.e. the leaves do not change their positions once they fall in the river.
For example, you are given integer X = 5 and array A such that:
A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4In second 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.
Write a function:
function solution($X, $A);
that, given a non-empty array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.
If the frog is never able to jump to the other side of the river, the function should return −1.
For example, given X = 5 and array A such that:
A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4the function should return 6, as explained above.
Write an efficient algorithm for the following assumptions:
- N and X are integers within the range [1..100,000];
- each element of array A is an integer within the range [1..X].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used PHP
Time spent on task 18 minutes
Notes
not defined yet
Code: 23:08:30 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
$maxk = 0; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
if ($maxk) return $maxk;
return -1; # reaching here means we didn't get to our position
}
User test case 1:
[1, []]
Analysis
Code: 23:08:50 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
$maxk = 0; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
if ($maxk) return $maxk;
return -1; # reaching here means we didn't get to our position
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
Analysis
Code: 23:21:19 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
print_r($inplace);
$maxk = 0; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
if ($maxk) return $maxk;
return -1; # reaching here means we didn't get to our position
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
Analysis
expand all
Example tests
1.
0.004 s
OK
stdout:
Array
(
[1] => 0
[3] => 1
[4] => 3
[2] => 4
[5] => 6
)
expand all
User tests
1.
0.004 s
OK
function result: -1
function result: -1
stdout:
Array
(
[1] => 0
)
1.
0.004 s
OK
function result: 1
function result: 1
stdout:
Array
(
[1] => 1
)
Code: 23:21:57 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
print_r($inplace);
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
Analysis
expand all
Example tests
1.
0.004 s
OK
stdout:
Array
(
[1] => 0
[3] => 1
[4] => 3
[2] => 4
[5] => 6
)
expand all
User tests
1.
0.004 s
OK
function result: 0
function result: 0
stdout:
Array
(
[1] => 0
)
1.
0.004 s
OK
function result: 1
function result: 1
stdout:
Array
(
[1] => 1
)
Code: 23:22:14 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
print_r($inplace);
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
User test case 3:
[5, [2,1])
Analysis
expand all
Example tests
1.
0.004 s
OK
stdout:
Array
(
[1] => 0
[3] => 1
[4] => 3
[2] => 4
[5] => 6
)
expand all
User tests
1.
0.004 s
OK
function result: 0
function result: 0
stdout:
Array
(
[1] => 0
)
1.
0.004 s
OK
function result: 1
function result: 1
stdout:
Array
(
[1] => 1
)
1.
0.001 s
RUNTIME ERROR,
invalid input, unexpected token at the end of input: ')'
Code: 23:23:17 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
print_r($inplace);
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
User test case 3:
[5, [2, 1]]
User test case 4:
[5, [2,1,5])
Analysis
expand all
Example tests
1.
0.004 s
OK
stdout:
Array
(
[1] => 0
[3] => 1
[4] => 3
[2] => 4
[5] => 6
)
expand all
User tests
1.
0.004 s
OK
function result: 0
function result: 0
stdout:
Array
(
[1] => 0
)
1.
0.004 s
OK
function result: 1
function result: 1
stdout:
Array
(
[1] => 1
)
1.
0.004 s
OK
function result: -1
function result: -1
stdout:
Array
(
[2] => 0
[1] => 1
)
1.
0.001 s
RUNTIME ERROR,
invalid input, unexpected token at the end of input: ')'
Code: 23:23:34 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
print_r($inplace);
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
User test case 3:
[5, [2, 1]]
User test case 4:
[5, [2, 1, 5]]
Analysis
expand all
Example tests
1.
0.004 s
OK
stdout:
Array
(
[1] => 0
[3] => 1
[4] => 3
[2] => 4
[5] => 6
)
expand all
User tests
1.
0.004 s
OK
function result: 0
function result: 0
stdout:
Array
(
[1] => 0
)
1.
0.004 s
OK
function result: 1
function result: 1
stdout:
Array
(
[1] => 1
)
1.
0.004 s
OK
function result: -1
function result: -1
stdout:
Array
(
[2] => 0
[1] => 1
)
1.
0.004 s
OK
function result: -1
function result: -1
stdout:
Array
(
[2] => 0
[1] => 1
[5] => 2
)
Code: 23:23:53 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
User test case 3:
[5, [2, 1]]
User test case 4:
[5, [2, 1, 5]]
Analysis
expand all
User tests
1.
0.008 s
OK
function result: 0
function result: 0
1.
0.008 s
OK
function result: 1
function result: 1
1.
0.008 s
OK
function result: -1
function result: -1
1.
0.008 s
OK
function result: -1
function result: -1
Code: 23:25:13 UTC,
php,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}
User test case 1:
[1, [1]]
User test case 2:
[1, [2, 1]]
User test case 3:
[5, [2, 1]]
User test case 4:
[5, [2, 1, 5]]
Analysis
expand all
User tests
1.
0.004 s
OK
function result: 0
function result: 0
1.
0.004 s
OK
function result: 1
function result: 1
1.
0.004 s
OK
function result: -1
function result: -1
1.
0.004 s
OK
function result: -1
function result: -1
Code: 23:25:15 UTC,
php,
final,
score:
100
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($X, $A) {
// write your code in PHP7.0
$inplace = []; # positions in place to our goal, space O(X). Index by position number 0..$X
foreach ($A as $k => $pos) {
# time O(N) - array of N integers
if ($pos <= $X) { # we are only interested in positions wuthin $X
# count positions - but store the first $k key seconds which this position is reached.
# We are not interested in when the second leaf of this same position falls.
if (!isset($inplace[$pos])) $inplace[$pos] = $k;
}
}
$maxk = -1; # max key value which is the longest time for the needed leaf to fall
for ($i=1; $i <= $X; $i++) {
# go through every position
if (isset($inplace[$i])) {
$tempk = $inplace[$i]; //k seconds for this leaf to fall
$maxk = ($tempk > $maxk) ? $tempk : $maxk;
}
else return -1; # if this position is not set, the leaf does not fall, so we exit
}
return $maxk;
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
expand all
Performance tests
1.
0.008 s
OK
2.
0.008 s
OK
1.
0.004 s
OK
1.
0.032 s
OK
2.
0.028 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.016 s
OK