Amazon Interview Question for SDE1s

Team: Machine learning
Country: India
Interview Type: Phone Interview

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0
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I don't understand the question. If it's simply the sum of the left == sum of the right, then no split would make your example of [-1, 100, 1, 98, 1] work. That question would actually be fairly difficult to do and would require multiple passes.

If the question is at what point does some left slice and some right slice have the same sum, then simply have a left sum pointer and a right sum pointer.

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0
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``````private static int FindIndexWhereLeftSumEqualRightSum(int[] arr)
{
if (arr == null || arr.Length == 0)
return -1;
int i = 1;
int j = arr.Length - 2;
int leftSum = arr[0];
int rightSum = arr[arr.Length-1];
while (i < j)
{
if (leftSum < rightSum)
{
leftSum += arr[i];
i++;
}
else
{
rightSum += arr[j];
j--;
}
}

if (leftSum == rightSum)
{
return leftSum;
}
return -1;
}``````

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0
of 0 vote

Java
Complexity: O(n)

``````public class SumOfLeftEqualsSumofRight {

public static void main(String[] args) {
int arr[] = {-1, 100, 1, 98, 1};
SumOfLeftEqualsSumofRight sumOfLeftEqualsSumofRight = new SumOfLeftEqualsSumofRight();
int totalSum = getSum(arr);
System.out.println(sumOfLeftEqualsSumofRight.getElementWithEqualSum(arr,totalSum));
}

public static int getSum (int []arr){
int sum=0;
for(int curr:arr){
sum=sum+curr;
}
return sum;
}

public int getElementWithEqualSum(int arr[], int sum){
if (arr == null || arr.length == 0)
return -1;
int leftTotal = 0;//arr[0];
int rightTotal = sum-arr[0];
for(int i=1;i<arr.length;i++){
leftTotal = leftTotal + arr[i-1];
rightTotal = rightTotal - arr[i];
if(leftTotal==rightTotal){
return i;
}
}
return -1;
}
}``````

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0
of 0 vote

it is a phone interview question and should not complicate by aping solutions of similar type of problem

``````int findIndexHalfSum(int a[], int sz)
{
int lsum = 0, rsum = 0;

for (int i = 1; i++ ; i<sz) rsum +=a[i];

for (int i = 1; i++ ; i<=sz-2)
{
lsum +=a[i-1]; rsum -= a[i];

if (lsum == rsum) return i;
}

return -1;
}``````

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