You are given integers K, M and a non-empty array A consisting of N integers. Every element of the array is not greater than M.
You should divide this array into K blocks of consecutive elements. The size of the block is any integer between 0 and N. Every element of the array should belong to some block.
The sum of the block from X to Y equals A[X] + A[X + 1] + ... + A[Y]. The sum of empty block equals 0.
The large sum is the maximal sum of any block.
For example, you are given integers K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 2The array can be divided, for example, into the following blocks:
- [2, 1, 5, 1, 2, 2, 2], [], [] with a large sum of 15;
- [2], [1, 5, 1, 2], [2, 2] with a large sum of 9;
- [2, 1, 5], [], [1, 2, 2, 2] with a large sum of 8;
- [2, 1], [5, 1], [2, 2, 2] with a large sum of 6.
The goal is to minimize the large sum. In the above example, 6 is the minimal large sum.
Write a function:
function solution($K, $M, $A);
that, given integers K, M and a non-empty array A consisting of N integers, returns the minimal large sum.
For example, given K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 2the function should return 6, as explained above.
Write an efficient algorithm for the following assumptions:
- N and K are integers within the range [1..100,000];
- M is an integer within the range [0..10,000];
- each element of array A is an integer within the range [0..M].
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($K, $M, $A) {
$min = 0;
$max = 0;
for ($i = 0; $i < count( $A ); $i++) {//get the sum as max, and the largest number as min
$max += $A[$i];
$min = max($min, $A[$i]);
}
$result = $max;
while ($min <= $max) {
$mid = intval( ($min + $max) / 2 );
if (divisionSolvable($mid, $K - 1, $A)) {
$max = $mid - 1;
$result = $mid;
} else {
$min = $mid + 1;
}
}
return $result;
}
function divisionSolvable($mid, $k, $a) {
$sum = 0;
for ($i = 0; $i < count( $a ); $i++) {
$sum += $a[$i];
if ($sum > $mid) {
$sum = $a[$i];
$k--;
}
if ($k < 0) {
return false;
}
}
return true;
}
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($K, $M, $A) {
$min = 0;
$max = 0;
for ($i = 0; $i < count( $A ); $i++) {//get the sum as max, and the largest number as min
$max += $A[$i];
$min = max($min, $A[$i]);
}
$result = $max;
while ($min <= $max) {
$mid = intval( ($min + $max) / 2 );
if (divisionSolvable($mid, $K - 1, $A)) {
$max = $mid - 1;
$result = $mid;
} else {
$min = $mid + 1;
}
}
return $result;
}
function divisionSolvable($mid, $k, $a) {
$sum = 0;
for ($i = 0; $i < count( $a ); $i++) {
$sum += $a[$i];
if ($sum > $mid) {
$sum = $a[$i];
$k--;
}
if ($k < 0) {
return false;
}
}
return true;
}
// you can write to stdout for debugging purposes, e.g.
// print "this is a debug message\n";
function solution($K, $M, $A) {
$min = 0;
$max = 0;
for ($i = 0; $i < count( $A ); $i++) {//get the sum as max, and the largest number as min
$max += $A[$i];
$min = max($min, $A[$i]);
}
$result = $max;
while ($min <= $max) {
$mid = intval( ($min + $max) / 2 );
if (divisionSolvable($mid, $K - 1, $A)) {
$max = $mid - 1;
$result = $mid;
} else {
$min = $mid + 1;
}
}
return $result;
}
function divisionSolvable($mid, $k, $a) {
$sum = 0;
for ($i = 0; $i < count( $a ); $i++) {
$sum += $a[$i];
if ($sum > $mid) {
$sum = $a[$i];
$k--;
}
if ($k < 0) {
return false;
}
}
return true;
}
The solution obtained perfect score.