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leetcode/problems/src/array/FindPivotIndex.java at master · KindleBooks66/leetcode · GitHub
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array
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FindPivotIndex.java
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FindPivotIndex.java
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package
array
;
import
java
.
util
.*;
/**
* Created by gouthamvidyapradhan on 06/08/2019 Given an array of integers nums, write a method that
* returns the "pivot" index of this array.
*
* <p>We define the pivot index as the index where the sum of the numbers to the left of the index
* is equal to the sum of the numbers to the right of the index.
*
* <p>If no such index exists, we should return -1. If there are multiple pivot indexes, you should
* return the left-most pivot index.
*
* <p>Example 1:
*
* <p>Input: nums = [1, 7, 3, 6, 5, 6] Output: 3 Explanation: The sum of the numbers to the left of
* index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3. Also, 3 is the
* first index where this occurs.
*
* <p>Example 2:
*
* <p>Input: nums = [1, 2, 3] Output: -1 Explanation: There is no index that satisfies the
* conditions in the problem statement.
*
* <p>Note:
*
* <p>The length of nums will be in the range [0, 10000]. Each element nums[i] will be an integer in
* the range [-1000, 1000].
*
* <p>Solution: O(N) maintain a prefix and posfix sum array and then use this to arrive at the
* answer.
*/
public
class
FindPivotIndex
{
public
static
void
main
(
String
[]
args
) {}
public
int
pivotIndex
(
int
[]
nums
) {
if
(
nums
.
length
==
1
)
return
0
;
int
[]
left
=
new
int
[
nums
.
length
];
int
[]
right
=
new
int
[
nums
.
length
];
left
[
0
] =
nums
[
0
];
for
(
int
i
=
1
;
i
<
nums
.
length
;
i
++) {
left
[
i
] =
left
[
i
-
1
] +
nums
[
i
];
}
right
[
nums
.
length
-
1
] =
nums
[
nums
.
length
-
1
];
for
(
int
i
=
nums
.
length
-
2
;
i
>=
0
;
i
--) {
right
[
i
] =
right
[
i
+
1
] +
nums
[
i
];
}
for
(
int
i
=
0
;
i
<
nums
.
length
;
i
++) {
int
l
,
r
;
if
(
i
==
0
) {
l
=
0
;
}
else
{
l
=
left
[
i
-
1
];
}
if
(
i
==
nums
.
length
-
1
) {
r
=
0
;
}
else
{
r
=
right
[
i
+
1
];
}
if
(
l
==
r
)
return
i
;
}
return
-
1
;
}
}
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