A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3We can split this tape in four places:
- P = 1, difference = |3 − 10| = 7
- P = 2, difference = |4 − 9| = 5
- P = 3, difference = |6 − 7| = 1
- P = 4, difference = |10 − 3| = 7
Write a function:
function solution(A);
that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3the function should return 1, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [2..100,000];
- each element of array A is an integer within the range [−1,000..1,000].
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p;
var index;
var leftSum;
var rightSum;
var totalSum=0;
var last_minimum = 100000;
if(A.length==2) return (Math.abs(A[0]-A[1]));
if(A.length==1) return (Math.abs(A[0]));
for(p=1; p <= A.length-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p;
var index;
var leftSum;
var rightSum;
var totalSum=0;
var last_minimum;
if(A.length==2) return (Math.abs(A[0]-A[1]));
if(A.length==1) return (Math.abs(A[0]));
for(p=1; p <= A.length-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
Invalid result type, integer expected, 'undefined' found Perhaps you are missing a 'return'?
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p;
var index;
var leftSum;
var rightSum;
var totalSum=0;
var last_minimum;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(p=1; p <= A.length-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
Invalid result type, integer expected, 'undefined' found Perhaps you are missing a 'return'?
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p;
var index;
var leftSum;
var rightSum;
var totalSum=0;
var last_minimum=10000;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(p=1; p <= A.length-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p=0;
var index=0;
var leftSum=0;
var rightSum=0;
var totalSum=0;
var last_minimum=10000;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(p=1; p <= A.length-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p=0;
var index=0;
var leftSum=0;
var rightSum=0;
var totalSum=0;
var N = A.length;
var last_minimum=100000;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(p=1; p <= N-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p=0;
var index=0;
var leftSum=0;
var rightSum=0;
var totalSum=0;
var N = A.length;
var last_minimum=100000;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(index=0; index < N; index++)
totalSum = totalSum + A[index];
for(p=1; p <= N-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p=0;
var index=0;
var leftSum=0;
var rightSum=0;
var totalSum=0;
var N = A.length;
var last_minimum=100000;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(index=0; index < N; index++)
totalSum = totalSum + A[index];
for(p=1; p <= N-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var p=0;
var index=0;
var leftSum=0;
var rightSum=0;
var totalSum=0;
var N = A.length;
var last_minimum=100000;
if(A.length == 2)
return (Math.abs(A[0]-A[1]));
if(A.length == 1)
return (Math.abs(A[0]));
for(index=0; index < N; index++)
totalSum = totalSum + A[index];
for(p=1; p <= N-1; p++){
leftSum += A[p - 1];
rightSum = totalSum - leftSum;
current_min = Math.abs(leftSum - rightSum);
last_minimum = current_min < last_minimum ? current_min : last_minimum;
if (last_minimum === 0)
break;
}
return last_minimum;
}
The solution obtained perfect score.