## Facebook Interview Question for SDE1s

• 0

Country: United States

Comment hidden because of low score. Click to expand.
2
of 2 vote

``````public static int maxSum(int[] nums, int k)
{
if (nums.length == 0) return 0;

int curSum = nums[0];
int maxSum = curSum;
int start = 0;

for (int end = 1; end < nums.length; end++)
{
if (curSum <= 0)
{
curSum = 0;
start = end;
}

if (end - start + 1 > k)
{
curSum -= nums[start];
start++;
}
//assume for the moment that curSum >= 0
//increment

curSum += nums[end];

if (curSum > maxSum) maxSum = curSum;

}

return maxSum;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

I have written below code and tested for several test cases working fine -

Time complexity =~ O(n)

``````public static int findMaxSum(int[] ar,int k){
int max = ar[0]; // max inititalize with first element
int l = ar.length; // Length of array
int sum = ar[0]; // sum included ith item
int sum_e = 0; // Excluded ith item
int j = 0; // starting point of k or < k items sum
int i = 1; // index pointer
while(i < l){
if(i-j < k){
sum_e = sum;
sum = sum + ar[i];
//Used temp to find max of sum and sum_e without changing existing value of sum and
// sum_e as these will be needed for next iteration
int temp  = sum > sum_e ? sum : sum_e;
max = max > temp ? max : temp;

i++;
}
if(ar[j] < 0 || i-j >= k) {
sum = sum -ar[j];
j++;
max = max > sum ? max : sum;
}

}

return max;
}

public static void main(String[] args) {
// TODO Auto-generated method stub

int[] ar0 = {-5,-10};
int[] ar1 = {-3,10};
int[] ar2 = {3,-10};
int[] ar3 = {-2,5,60,-10,23};
int[] ar4 = {6,2,4,-1,5,9,-2,10};

System.out.println(findMaxSum(ar0, 4));
System.out.println(findMaxSum(ar1, 4));
System.out.println(findMaxSum(ar2, 4));
System.out.println(findMaxSum(ar3, 4));
System.out.println(findMaxSum(ar4, 4));
}``````

o/p -
-5
10
3
78
22

Thanks

Comment hidden because of low score. Click to expand.
0
of 0 vote

Concise version in Scala:

``````object MaxWindowKSum {

def main(args: Array[String]): Unit = {
val a = Array(10, 12, 15, 18, 20, 4, 5, 9, 28)
println(getMaxWindowKSum(a, 3))
}

def getMaxWindowKSum(a:Array[Int], k:Int) : Int = {
var max = 0
var i = 0

while (i + k < a.length) {
val sum =  a.slice(i , i + k).sum
if (max < sum) max = sum
i += 1
}
return max
}
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````int MaxSubarray(vector<int> const &a, int k)
{
int max_sum = numeric_limits<int>::min();
if (k > 0) {
int start = 0;
int sum = 0;
for (int i = 0; i < a.size(); ++i) {
while (i - start + 1 > k ||
(start < i && a[start] <= 0))
{
sum -= a[start];
++start;
}
sum += a[i];
max_sum = max(max_sum, sum);
if (sum <= 0) {
sum = 0;
start = i + 1;
}
}
}
return max_sum;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````//Time: O(N), Space: O(N)
public int maxSum(int[] arr, int k){
for(int i = 1; i < arr.length; i++){
arr[i] += arr[i - 1];
}
int maxSum = arr[0];
for(int i = 1; i < k; i++){
maxSum = Math.max(maxSum, Math.max(arr[i], arr[i] - arr[q.peekFirst()]);
while(!q.isEmpty()){
if(arr[i] < arr[q.peekFirst()]){
q.pollFirst();
}
}

}
for(int i = k; i < arr.length; i++){
while(q.peekFirst() < i - k){
q.pollFirst();
}
maxSum = Math.max(maxSum,arr[i] - arr[q.peekFirst()]);
while(!q.isEmpty()){
if(arr[i] < arr[q.peekFirst()]){
q.pollFirst();
}
}
}
return maxSum;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Keep an index of the start of the segment when length of segments get above k, remove start element from sum and add new element, in this way the length of segment will always be <= k, and new elements can be checked for maximum sum.

{{

int start = 0;
int maxsum = -infinity;

for(int i = 0; i < array.length; i++)
{
if(start - i > k)
{
sum = sum - array[start];
start++;
}

sum += array[i];

if(sum > maxsum)
maxsum = sum;
if(sum < 0)
{
start = i + 1;
sum = 0;
}
}

}}}

Comment hidden because of low score. Click to expand.
0
of 0 vote

Keep an index of the start of the segment when length of segments get above k, remove start element from sum and add new element, in this way the length of segment will always be <= k, and new elements can be checked for maximum sum.

``````int start = 0;
int maxsum = -infinity;

for(int i = 0; i < array.length; i++)
{
if(start - i > k)
{
sum = sum - array[start];
start++;
}

sum += array[i];

if(sum > maxsum)
maxsum = sum;
if(sum < 0)
{
start = i + 1;
sum = 0;
}
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Time: O(n)
Space O(1)

``````public int maxSum(final int[] input) {
Preconditions.checkNotNull(input, "Illegal input!");
int maxSum = Integer.MIN_VALUE;
int currentSum = 0;
for (int i = 0; i < input.length; ++i) {
currentSum = Math.max(currentSum + input[i], input[i]);
maxSum = Math.max(currentSum, maxSum);
}
return maxSum;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

O(n) solution:

``````public static int maxSumWithK() {
int[] a = { 1, 1, 1, 1, 1, 1, 1 };
int k = 2;
int maxSum[] = new int[a.length];
int sum = a[0];
maxSum[0] = sum;
for (int i = 1; i < maxSum.length; i++) {
sum += a[i];
if (sum > a[i]) {
maxSum[i] = sum;
} else {
maxSum[i] = a[i];
sum = a[i];
}
}

sum = 0;
for (int i = 0; i < k; i++) {
sum += a[i];
}
int max = sum;
for (int i = k; i < maxSum.length; i++) {
sum = sum + a[i] - a[i - k];
if (sum + maxSum[i - k] > max)
max = sum + maxSum[i - k];
}
return max;``````

}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````def maxSum(self, s,k):
begin=0
sum=0
max_sum=0
for i in xrange(len(s)):
if i-start+1 >k :
sum = sum - s[i]
start+=1
sum = sum + s[i]
if s[i] > sum :
sum = s[i]
start = i
if sum > max_sum:
max_sum = sum
return max_sum``````

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