LeetCode – Maximum Subarray (Java)
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray [4,−1,2,1] has the largest sum = 6.
1. Dynamic Programming Solution
The changing condition for dynamic programming is "We should ignore the sum of the previous n-1 elements if nth element is greater than the sum."
public class Solution { public int maxSubArray(int[] A) { int max = A[0]; int[] sum = new int[A.length]; sum[0] = A[0]; for (int i = 1; i < A.length; i++) { sum[i] = Math.max(A[i], sum[i - 1] + A[i]); max = Math.max(max, sum[i]); } return max; } } |
1. Simple Solution
Mehdi provided the following solution in his comment. The time space is optimized to be O(1).
public int maxSubArray(int[] A) { int newsum=A[0]; int max=A[0]; for(int i=1;i<A.length;i++){ newsum=Math.max(newsum+A[i],A[i]); max= Math.max(max, newsum); } return max; } |
This problem is asked by Palantir.
<pre><code> String foo = "bar"; </code></pre>
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Why does the so-called “Naive Solution” not work? I think the algorithm is correct, and it passes the OJ as well.
原理就是错的
because it’s a wrong method
Hi, what about the actual subarray that should be returned? Only the max sum is returned in the two solutions above.
you can also do it with O(1) space complexity instead of O(n):
public int maxSubArray(int[] A) { int newsum=A[0]; int max=A[0]; for(int i=1;i<A.length;i++){ newsum=Math.max(newsum+A[i],A[i]); max= Math.max(max, newsum); } return max; }Native Solution is wrong, but you can change some row, and it will be ok!
like this:
public static int maxSubArray(int[] A) {
int sum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < A.length; i++) {
sum += A[i];
if (sum < 0)
sum = A[i]; // change
maxSum = Math.max(maxSum, sum); // change
}
return maxSum;
}
Based on the second method(DP style), I write this for printing the sub array in details. I tested the example in the question and it works.
public int maxSubArray(int[] A) {
int max = A[0];
int[] sum = new int[A.length];
sum[0] = A[0];
int begin = 0;
int end = 0;
for(int i=1; i sum[i-1] + A[i]) {
sum[i] = A[i];
newBegin = i;
newEnd = i;
} else {
sum[i] = sum[i-1] + A[i];
newEnd = i;
}
if(sum[i] > max) {
max = sum[i];
begin = newBegin;
end = newEnd;
}
}
// print the subArray
for(int i = begin; i <= end; i++) {
System.out.print(A[i]+ " ");
}
System.out.println();
return max;
}
The “Naive Solution” works, nothing’s wrong with that line, I think it’s not working anymore if you change that row.
An array with all negative number will break naive solution, since “(containing at least one number) “
Will something like this work? We observe that the if the element about to be added to create the next summation is already greater than the previous summation, then we will ignore the previous summation and start a new maxSum from the current index. The code would look something like:
[code]
public int maxSum(int[] arr) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0, sumSoFar = -1, sum; i sum) {
sumSoFar = arr[i];
} else {
sumSoFar = sum;
}
maxSum = Math.max(maxSum, sumSoFar);
}
return maxSum;
}
// Test input and iterations
[−2,1,−3,4,−1,2,1,−5,4]
i = 0; sum = -3, sumSoFar = -2, maxSum = -2 ;
i = 1; sum = -1, sumSoFar = -1, maxSum = -1 ;
i = 2; sum = -4, sumSoFar = -3, maxSum = -3 ;
i = 3; sum = 1, sumSoFar = 4, maxSum = 4 ;
i = 4; sum = 3, sumSoFar = 3, maxSum = 3 ;
i = 5; sum = 5, sumSoFar = 5, maxSum = 5 ;
i = 6; sum = 6, sumSoFar = 6, maxSum = 6 ;
i = 7; sum = 1, sumSoFar = 1, maxSum = 6 ;
i = 8; sum = 5, sumSoFar = 5, maxSum = 6 ;
[/code]
I have added this solution, thanks!
I think your solution is also correct.
This solution is wrong, I tried it using { -1, -2, 5, -2 }.
Thanks! I was wondering how you guys created code like posting.
Put code between <pre> and </pre>
The first solution is correct. It’s same to the last solution
Above solution is not correct. Pls find the rectified complete solution below:
#include
using namespace std;
int max(int a, int b)
{
if(a>b)
return a;
return b;
}
int maxSubArray(int* A, int len) {
int sum = 0;
int maxSum = -1000000; //some very large negative number
for (int i = 0; i < len; i++) {
if (sum < 0){
sum = A[i]; // change
}
else
{
sum += A[i];
}
maxSum = max(maxSum, sum); // change
}
return maxSum;
}
int main() {
int a[] = {-1,-2,3,4,-5,6};
cout<<maxSubArray(a, 6);
return 0;
}
The DP solution is not correct. The condition should be “we ignore the previous n-1 sum if it were less than or equal to 0”.
The naive solution should work on {-1, -2, 5, -2}.
could you double check plz?
Check this out O(1) with position:
public static int maxSubArray(int[] A) {
int newsum=A[0];
int max=A[0];
int j = 0;
int i_max = 0;
for(int i=1; i<A.length; i++){
newsum += A[i];
if (newsum max) {
max = newsum;
i_max = i;
}
}
System.out.println(j + “,” + i_max + ” = ” + max);
return max;
}
Copy pasted from eclipse ide.I m not sure why it appears this way. Sorry for that
Solution with array indexes:
public static void maxArray(int[] arr){
int prevStart = -1;
int currentStart = prevStart;
int maxHere = 0;
int maxSoFar = 0;
int end = -1;
for(int i = 0; i< arr.length; i++){
if(arr[i] < 0 && maxSoFar == 0){
continue;
}
maxHere += arr[i];
if(currentStart == -1){
currentStart = i;
}
if(maxHere maxSoFar){
maxSoFar = maxHere;
prevStart = currentStart;
end = i;
}
}
System.out.println(“Max Sum:” + maxSoFar);
System.out.println(“Start: ” + prevStart);
System.out.println(“End: ” + end);
}
I will still work. plz test yourself.
Extending Mehdi’s solution, you can track the subarray (indices) using 2 more variables startIndex and endIndex (initialize before the for loop and add below code inside the for loop at the end):
if (max == A[i])
{
startIndex = i;
}
else if (max == newsum)
{
endIndex = i;
}
only fails with all negative numbers
Guess it fails for lot of cases, For Ex input Array:
5, -4, 9, -3500, 25, -3, 12, 5900, 15, -23, 18, 14, 11, 93, 12, 34, 56, -125, 66, -255, 95, 68, 4558, -2, 203
/* Here is the solution using kadane’s algorithm but it only works when there is at least one positive element in the given array. */
public static int maxSubArray(int[] array) {
int maxSum = 0, newSum = 0;
for(int i = 0; i 0) {
newSum = newSum + array[i];
} else {
newSum = 0;
}
maxSum = Math.max(maxSum, newSum);
}
return maxSum;
}
This seems simpler:
public static int maxSubArray(int[] a) {
int r = 0;
int l = 0;
int rsum = 0;
int lsum = 0;
int rmax = Integer.MIN_VALUE;
int lmax = Integer.MIN_VALUE;
for (int i = 0; i rmax) {
rmax = rsum;
r = i;
}
if (lsum > lmax) {
lmax = lsum;
l = a.length - i - 1;
}
}
int rtn = rmax + lmax - rsum;
System.out.println("a[" + l + "] + ... + a[" + r + "] = " + rtn);
return rtn;
}
Could you please explain why the complexity is O(1) and not O(n)?
As I see it, the for loop is repeated n-1 times.
much clean solution would be
// input the position of element of array “a”
int maxSub(int pos){
if(pos == 0)
return a[0];
else{
return Math.max(maxSub(pos-1)+a[pos], a[pos]);
}
}