left+(right-left)/2

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0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 1/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

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0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 5/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 1/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 1/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

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0

int mid = (left + right)/2;

Popularity 3/10 Helpfulness 1/10 Language whatever

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 2/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 1/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

//why left+(right-left)/2 will not overflow?

//Suppose (to make the example easier) the maximum integer is 100, 
//left = 50, and right = 80. If you use the naive formula:

int mid = (left + right)/2;
//the addition will result in 130, which overflows.

//If you instead do:

int mid = left + (right - left)/2;
//you can't overflow in (right - left) because you're subtracting 
//a smaller number from a larger number. That always results in
//an even smaller number, so it can't possibly go over the maximum. 
//E.g. 80 - 50 = 30.

//And since the result is the average of left and right, 
//it must be between them. Since these are both less than the maximum
//integer, anything between them is also less than the maximum,
//so there's no overflow.

Popularity 2/10 Helpfulness 1/10 Language java

Source: stackoverflow.com

Tags: java

Share

Link to this answer
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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

int mid = (left + right)/2;

Popularity 1/10 Helpfulness 1/10 Language whatever

Source: stackoverflow.com

Tags: whatever

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

Comment

0

int mid = (left + right)/2;

Popularity 2/10 Helpfulness 1/10 Language whatever

Source: stackoverflow.com

Tags: whatever

Share

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Contributed on Jul 26 2022

freelance developer

0 Answers Avg Quality 2/10

left+(right-left)/2

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